301. A compound contains \(24\,g\) of carbon and \(4\,g\) of hydrogen only. Using \(\mathrm{C}=12\) and \(\mathrm{H}=1\), its empirical formula is:
ⓐ. \(\mathrm{CH}\)
ⓑ. \(\mathrm{CH_2}\)
ⓒ. \(\mathrm{C_2H_4}\)
ⓓ. \(\mathrm{C_4H_2}\)
Correct Answer: \(\mathrm{CH_2}\)
Explanation: \( \textbf{Given masses:} \) carbon \(=24\,g\), hydrogen \(=4\,g\).
\( \textbf{Convert carbon mass to moles:} \)
\[
n_{\mathrm{C}}=\frac{24\,g}{12\,g\,mol^{-1}}=2\,mol
\]
\( \textbf{Convert hydrogen mass to moles:} \)
\[
n_{\mathrm{H}}=\frac{4\,g}{1\,g\,mol^{-1}}=4\,mol
\]
\( \textbf{Mole ratio:} \)
\[
\mathrm{C}:\mathrm{H}=2:4
\]
\( \textbf{Simplest whole-number ratio:} \)
\[
2:4=1:2
\]
\( \textbf{Final answer:} \) the empirical formula is \(\mathrm{CH_2}\).
The formula \(\mathrm{C_2H_4}\) has the same ratio but is not reduced to the simplest empirical formula.
302. A compound has empirical formula \(\mathrm{CH_2O}\) and molar mass \(180\,g\,mol^{-1}\). The molecular formula is:
ⓐ. \(\mathrm{CH_2O}\)
ⓑ. \(\mathrm{C_2H_4O_2}\)
ⓒ. \(\mathrm{C_6H_{12}O_6}\)
ⓓ. \(\mathrm{C_{12}H_{24}O_{12}}\)
Correct Answer: \(\mathrm{C_6H_{12}O_6}\)
Explanation: \( \textbf{Empirical formula:} \) \(\mathrm{CH_2O}\).
\( \textbf{Empirical formula mass:} \)
\[
12+2(1)+16=30\,g\,mol^{-1}
\]
\( \textbf{Given molar mass:} \) \(180\,g\,mol^{-1}\).
\( \textbf{Find multiplier:} \)
\[
\text{multiplier}=\frac{180}{30}=6
\]
\( \textbf{Multiply empirical subscripts by \(6\):} \)
\[
\mathrm{C_1H_2O_1}\times6=\mathrm{C_6H_{12}O_6}
\]
\( \textbf{Final answer:} \) the molecular formula is \(\mathrm{C_6H_{12}O_6}\).
The molecular formula must have a molar mass matching the given value, not only the same percentage composition.
303. A compound has empirical formula \(\mathrm{NO_2}\) and molecular mass \(92\,u\). The molecular formula is:
ⓐ. \(\mathrm{NO_2}\)
ⓑ. \(\mathrm{N_2O_4}\)
ⓒ. \(\mathrm{N_3O_6}\)
ⓓ. \(\mathrm{NO}\)
Correct Answer: \(\mathrm{N_2O_4}\)
Explanation: \( \textbf{Empirical formula:} \) \(\mathrm{NO_2}\).
\( \textbf{Use atomic masses:} \) \(\mathrm{N}=14\), \(\mathrm{O}=16\).
\( \textbf{Empirical formula mass:} \)
\[
14+2(16)=46\,u
\]
\( \textbf{Given molecular mass:} \) \(92\,u\).
\( \textbf{Multiplier:} \)
\[
\frac{92}{46}=2
\]
\( \textbf{Multiply subscripts:} \)
\[
\mathrm{NO_2}\times2=\mathrm{N_2O_4}
\]
\( \textbf{Final answer:} \) the molecular formula is \(\mathrm{N_2O_4}\).
The empirical formula is the simplest ratio, while the molecular formula gives the actual atom count in one molecule.
304. A student obtains the mole ratio \(\mathrm{C}:\mathrm{H}:\mathrm{O}=1.5:3:1\) while finding an empirical formula. The best next step is:
ⓐ. Use \(\mathrm{C_{1.5}H_3O}\) as the empirical formula
ⓑ. Round \(1.5\) down to \(1\) and write \(\mathrm{CH_3O}\)
ⓒ. Ignore oxygen because its ratio is smallest
ⓓ. Multiply ratios by \(2\) to get \(\mathrm{C_3H_6O_2}\)
Correct Answer: Multiply ratios by \(2\) to get \(\mathrm{C_3H_6O_2}\)
Explanation: Empirical formula subscripts must be whole numbers. The ratio \(1.5:3:1\) contains a half value, so it should not be written directly as a formula. Multiplying every term by \(2\) gives \(3:6:2\). This produces the empirical formula \(\mathrm{C_3H_6O_2}\). Rounding only one value changes the actual ratio and gives a formula not supported by the mole data.
305. A compound contains \(52.2\%\) carbon, \(13.0\%\) hydrogen, and \(34.8\%\) oxygen by mass. The empirical formula is:
ⓐ. \(\mathrm{CH_3O}\)
ⓑ. \(\mathrm{C_2H_6O}\)
ⓒ. \(\mathrm{C_3H_8O}\)
ⓓ. \(\mathrm{C_4H_{12}O_2}\)
Correct Answer: \(\mathrm{C_2H_6O}\)
Explanation: \( \textbf{Assume \(100\,g\) sample:} \) \(\mathrm{C}=52.2\,g\), \(\mathrm{H}=13.0\,g\), \(\mathrm{O}=34.8\,g\).
\( \textbf{Atomic masses:} \) \(\mathrm{C}=12\), \(\mathrm{H}=1\), \(\mathrm{O}=16\).
\( \textbf{Moles of carbon:} \)
\[
\frac{52.2}{12}=4.35\,mol
\]
\( \textbf{Moles of hydrogen:} \)
\[
\frac{13.0}{1}=13.0\,mol
\]
\( \textbf{Moles of oxygen:} \)
\[
\frac{34.8}{16}=2.175\,mol
\]
\( \textbf{Divide by smallest value \(2.175\):} \)
\[
\mathrm{C}:\mathrm{H}:\mathrm{O}=2:6:1
\]
\( \textbf{Final answer:} \) the empirical formula is \(\mathrm{C_2H_6O}\).
The ratio is already in whole numbers after division, so no further multiplier is needed.
306. Use the information below.
A compound has percentage composition \(\mathrm{C}=80.0\%\) and \(\mathrm{H}=20.0\%\). Its molar mass is \(30\,g\,mol^{-1}\). Use \(\mathrm{C}=12\) and \(\mathrm{H}=1\).
The molecular formula is:
ⓐ. \(\mathrm{CH_3}\)
ⓑ. \(\mathrm{C_2H_6}\)
ⓒ. \(\mathrm{C_3H_9}\)
ⓓ. \(\mathrm{C_2H_4}\)
Correct Answer: \(\mathrm{C_2H_6}\)
Explanation: \( \textbf{Assume \(100\,g\) sample:} \) carbon \(=80.0\,g\), hydrogen \(=20.0\,g\).
\( \textbf{Moles of carbon:} \)
\[
\frac{80.0}{12}=6.67\,mol
\]
\( \textbf{Moles of hydrogen:} \)
\[
\frac{20.0}{1}=20.0\,mol
\]
\( \textbf{Divide by smaller value \(6.67\):} \)
\[
\mathrm{C}:\mathrm{H}=1:3
\]
\( \textbf{Empirical formula:} \) \(\mathrm{CH_3}\).
\( \textbf{Empirical formula mass:} \)
\[
12+3(1)=15\,g\,mol^{-1}
\]
\( \textbf{Multiplier:} \)
\[
\frac{30}{15}=2
\]
\( \textbf{Final answer:} \) molecular formula \(=\mathrm{C_2H_6}\).
The empirical formula gives the simplest ratio, while the molar mass decides the actual molecular formula.
307. Two compounds have the same empirical formula \(\mathrm{CH_2}\). Compound P has molar mass \(28\,g\,mol^{-1}\), and compound Q has molar mass \(56\,g\,mol^{-1}\). The correct molecular formulae of P and Q are:
ⓐ. P-\(\mathrm{CH_2}\), Q-\(\mathrm{CH_2}\)
ⓑ. P-\(\mathrm{C_2H_2}\), Q-\(\mathrm{C_4H_4}\)
ⓒ. P-\(\mathrm{C_4H_8}\), Q-\(\mathrm{C_2H_4}\)
ⓓ. P-\(\mathrm{C_2H_4}\), Q-\(\mathrm{C_4H_8}\)
Correct Answer: P-\(\mathrm{C_2H_4}\), Q-\(\mathrm{C_4H_8}\)
Explanation: \( \textbf{Empirical formula:} \) \(\mathrm{CH_2}\).
\( \textbf{Empirical formula mass:} \)
\[
12+2(1)=14\,g\,mol^{-1}
\]
\( \textbf{For compound P:} \)
\[
\frac{28}{14}=2
\]
\( \textbf{Molecular formula of P:} \)
\[
\mathrm{C_2H_4}
\]
\( \textbf{For compound Q:} \)
\[
\frac{56}{14}=4
\]
\( \textbf{Molecular formula of Q:} \)
\[
\mathrm{C_4H_8}
\]
\( \textbf{Final answer:} \) P is \(\mathrm{C_2H_4}\), and Q is \(\mathrm{C_4H_8}\).
The same empirical formula can correspond to different molecular formulae when molar masses are different.
308. A compound contains only nitrogen and oxygen. A \(4.60\,g\) sample contains \(1.40\,g\) nitrogen and \(3.20\,g\) oxygen. Using \(\mathrm{N}=14\) and \(\mathrm{O}=16\), the empirical formula is:
ⓐ. \(\mathrm{NO}\)
ⓑ. \(\mathrm{NO_2}\)
ⓒ. \(\mathrm{N_2O}\)
ⓓ. \(\mathrm{N_2O_5}\)
Correct Answer: \(\mathrm{NO_2}\)
Explanation: \( \textbf{Given masses:} \) nitrogen \(=1.40\,g\), oxygen \(=3.20\,g\).
\( \textbf{Moles of nitrogen:} \)
\[
n_{\mathrm{N}}=\frac{1.40\,g}{14\,g\,mol^{-1}}=0.100\,mol
\]
\( \textbf{Moles of oxygen:} \)
\[
n_{\mathrm{O}}=\frac{3.20\,g}{16\,g\,mol^{-1}}=0.200\,mol
\]
\( \textbf{Mole ratio:} \)
\[
\mathrm{N}:\mathrm{O}=0.100:0.200
\]
\( \textbf{Divide by \(0.100\):} \)
\[
\mathrm{N}:\mathrm{O}=1:2
\]
\( \textbf{Final answer:} \) the empirical formula is \(\mathrm{NO_2}\).
The ratio must be based on moles, not directly on the mass ratio \(1.40:3.20\).
309. A claim says, "The empirical formula of \(\mathrm{C_6H_{12}O_6}\) is also \(\mathrm{C_6H_{12}O_6}\) because changing subscripts changes the compound." The best correction is:
ⓐ. The empirical formula is \(\mathrm{CH_2O}\), because empirical formula gives the simplest whole-number ratio
ⓑ. The empirical formula is \(\mathrm{C_3H_6O_3}\), because all subscripts must be divided by \(2\) only
ⓒ. The empirical formula is \(\mathrm{C_6H_{12}O_6}\), because molecular and empirical formulae are always identical
ⓓ. The empirical formula is \(\mathrm{CHO}\), because all subscripts are ignored
Correct Answer: The empirical formula is \(\mathrm{CH_2O}\), because empirical formula gives the simplest whole-number ratio
Explanation: The molecular formula \(\mathrm{C_6H_{12}O_6}\) gives the actual number of atoms in one molecule. The empirical formula gives only the simplest whole-number ratio of atoms. Dividing the subscripts \(6:12:6\) by \(6\) gives \(1:2:1\). Therefore, the empirical formula is \(\mathrm{CH_2O}\). Empirical formula reduction does not say the molecule itself has fewer atoms; it reports the simplest composition ratio.
310. A graph is described below.
For a compound containing only X and Y, the x-axis shows moles of X and the y-axis shows moles of Y calculated from different samples. The plotted points lie on a straight line through the origin with slope \(2\).
The empirical formula is most likely:
ⓐ. \(\mathrm{XY}\)
ⓑ. \(\mathrm{XY_2}\)
ⓒ. \(\mathrm{X_2Y}\)
ⓓ. \(\mathrm{X_2Y_2}\)
Correct Answer: \(\mathrm{XY_2}\)
Explanation: \( \textbf{Graph axes:} \) x-axis \(=\) moles of X, y-axis \(=\) moles of Y.
\( \textbf{Slope meaning:} \)
\[
\text{slope}=\frac{\text{moles of Y}}{\text{moles of X}}
\]
\( \textbf{Given slope:} \) \(2\).
\( \textbf{Ratio from slope:} \)
\[
\frac{n_{\mathrm{Y}}}{n_{\mathrm{X}}}=2
\]
\( \textbf{Mole ratio:} \)
\[
\mathrm{X}:\mathrm{Y}=1:2
\]
\( \textbf{Final answer:} \) the empirical formula is \(\mathrm{XY_2}\).
The slope gives the relative mole amount of Y per mole of X, so it becomes the subscript ratio after simplification.
311. A sample of a compound gives \(0.250\,mol\) of X atoms and \(0.375\,mol\) of Y atoms. The empirical formula is:
ⓐ. \(\mathrm{XY}\)
ⓑ. \(\mathrm{X_2Y_3}\)
ⓒ. \(\mathrm{X_3Y_2}\)
ⓓ. \(\mathrm{XY_2}\)
Correct Answer: \(\mathrm{X_2Y_3}\)
Explanation: \( \textbf{Given mole amounts:} \) \(n_{\mathrm{X}}=0.250\,mol\), \(n_{\mathrm{Y}}=0.375\,mol\).
\( \textbf{Write ratio:} \)
\[
\mathrm{X}:\mathrm{Y}=0.250:0.375
\]
\( \textbf{Divide by the smaller value \(0.250\):} \)
\[
\mathrm{X}:\mathrm{Y}=1:1.5
\]
\( \textbf{Remove the half ratio by multiplying by \(2\):} \)
\[
1:1.5=2:3
\]
\( \textbf{Final answer:} \) the empirical formula is \(\mathrm{X_2Y_3}\).
The intermediate ratio \(1:1.5\) is not acceptable as a final formula because formula subscripts must be whole numbers.
312. A compound has empirical formula \(\mathrm{C_2H_5}\). Its vapour-density-based molar mass is found to be \(58\,g\,mol^{-1}\). Using \(\mathrm{C}=12\) and \(\mathrm{H}=1\), the molecular formula is:
ⓐ. \(\mathrm{C_2H_5}\)
ⓑ. \(\mathrm{C_6H_{15}}\)
ⓒ. \(\mathrm{CH_3}\)
ⓓ. \(\mathrm{C_4H_{10}}\)
Correct Answer: \(\mathrm{C_4H_{10}}\)
Explanation: \( \textbf{Empirical formula:} \) \(\mathrm{C_2H_5}\).
\( \textbf{Empirical formula mass:} \)
\[
2(12)+5(1)=29\,g\,mol^{-1}
\]
\( \textbf{Given molar mass:} \) \(58\,g\,mol^{-1}\).
\( \textbf{Find multiplier:} \)
\[
\frac{58}{29}=2
\]
\( \textbf{Multiply empirical subscripts by \(2\):} \)
\[
\mathrm{C_2H_5}\times2=\mathrm{C_4H_{10}}
\]
\( \textbf{Final answer:} \) the molecular formula is \(\mathrm{C_4H_{10}}\).
The empirical formula cannot be used as the molecular formula unless its formula mass matches the given molar mass.
313. A compound contains \(69.6\%\) oxygen and \(30.4\%\) nitrogen by mass. Using \(\mathrm{N}=14\) and \(\mathrm{O}=16\), its empirical formula is:
ⓐ. \(\mathrm{NO}\)
ⓑ. \(\mathrm{NO_2}\)
ⓒ. \(\mathrm{N_2O_3}\)
ⓓ. \(\mathrm{N_2O_5}\)
Correct Answer: \(\mathrm{N_2O_5}\)
Explanation: \( \textbf{Assume sample mass:} \) \(100\,g\).
\( \textbf{Mass of nitrogen:} \) \(30.4\,g\).
\( \textbf{Mass of oxygen:} \) \(69.6\,g\).
\( \textbf{Convert nitrogen to moles:} \)
\[
n_{\mathrm{N}}=\frac{30.4\,g}{14\,g\,mol^{-1}}=2.17\,mol
\]
\( \textbf{Convert oxygen to moles:} \)
\[
n_{\mathrm{O}}=\frac{69.6\,g}{16\,g\,mol^{-1}}=4.35\,mol
\]
\( \textbf{Divide by the smaller mole value:} \)
\[
\mathrm{N}:\mathrm{O}=\frac{2.17}{2.17}:\frac{4.35}{2.17}
\]
\[
\mathrm{N}:\mathrm{O}=1:2.00
\]
\( \textbf{First ratio:} \) \(\mathrm{NO_2}\).
\( \textbf{Check with exact percentage trend:} \) option \(\mathrm{NO_2}\) gives oxygen percentage \(\frac{32}{46}\times100=69.6\%\).
\( \textbf{Final answer:} \) the empirical formula is \(\mathrm{NO_2}\).
The mass percentages must be converted to moles before comparing atom ratios, because mass ratios are not directly subscript ratios.
314. A compound has empirical formula \(\mathrm{NO_2}\) and molecular mass \(92\,u\). Its molecular formula is:
ⓐ. \(\mathrm{NO_2}\)
ⓑ. \(\mathrm{N_2O_4}\)
ⓒ. \(\mathrm{N_2O_5}\)
ⓓ. \(\mathrm{NO}\)
Correct Answer: \(\mathrm{N_2O_4}\)
Explanation: \( \textbf{Empirical formula:} \) \(\mathrm{NO_2}\).
\( \textbf{Atomic masses:} \) \(\mathrm{N}=14\), \(\mathrm{O}=16\).
\( \textbf{Empirical formula mass:} \)
\[
14+2(16)=46\,u
\]
\( \textbf{Given molecular mass:} \) \(92\,u\).
\( \textbf{Find multiplier:} \)
\[
\frac{92}{46}=2
\]
\( \textbf{Multiply all empirical subscripts by \(2\):} \)
\[
\mathrm{NO_2}\times2=\mathrm{N_2O_4}
\]
\( \textbf{Final answer:} \) the molecular formula is \(\mathrm{N_2O_4}\).
The molecular formula keeps the same simplest ratio but uses the actual molecular mass to decide the true number of atoms.
315. A student obtains the mole ratio \(\mathrm{A}:\mathrm{B}=1:1.33\) while finding an empirical formula. The most suitable next step is:
ⓐ. Write \(\mathrm{AB_{1.33}}\) as the empirical formula
ⓑ. Multiply both by \(3\) to get \(\mathrm{A_3B_4}\)
ⓒ. Round \(1.33\) to \(1\) and write \(\mathrm{AB}\)
ⓓ. Multiply only B by \(3\) and write \(\mathrm{AB_4}\)
Correct Answer: Multiply both by \(3\) to get \(\mathrm{A_3B_4}\)
Explanation: Empirical formula subscripts must be whole numbers. The value \(1.33\) is close to \(\frac{4}{3}\), so the whole ratio \(1:1.33\) should be multiplied by \(3\). This gives approximately \(3:4\). Multiplying only one part of the ratio would change the relationship between the elements. Rounding \(1.33\) to \(1\) would also lose the composition information contained in the mole data.
316. Study the table below for empirical formula work.
| Step | Student action |
|---|
| P | Assumes \(100\,g\) sample from percentage data |
| Q | Converts each element's mass into moles |
| R | Divides each mole value by the largest mole value |
| S | Converts fractional ratios into whole-number ratios when needed |
The step that needs correction is:
ⓐ. Step P
ⓑ. Step Q
ⓒ. Step R
ⓓ. Step S
Correct Answer: Step R
Explanation: Assuming a \(100\,g\) sample is a useful method when percentage composition is given. Converting masses into moles is necessary because formula subscripts represent atom or mole ratios, not mass ratios. The mole values should be divided by the smallest mole value, not by the largest. Dividing by the smallest makes at least one ratio term equal to \(1\), which helps produce the simplest whole-number ratio. Fractional values such as \(1.5\) or \(1.33\) may then be removed by multiplying all ratio terms by a suitable integer.
317. A compound has empirical formula \(\mathrm{CH}\). Its molar mass is \(78\,g\,mol^{-1}\). The molecular formula is:
ⓐ. \(\mathrm{CH}\)
ⓑ. \(\mathrm{C_2H_2}\)
ⓒ. \(\mathrm{C_6H_6}\)
ⓓ. \(\mathrm{C_6H_{12}}\)
Correct Answer: \(\mathrm{C_6H_6}\)
Explanation: \( \textbf{Empirical formula:} \) \(\mathrm{CH}\).
\( \textbf{Empirical formula mass:} \)
\[
12+1=13\,g\,mol^{-1}
\]
\( \textbf{Given molar mass:} \) \(78\,g\,mol^{-1}\).
\( \textbf{Multiplier:} \)
\[
\frac{78}{13}=6
\]
\( \textbf{Multiply empirical subscripts by \(6\):} \)
\[
\mathrm{CH}\times6=\mathrm{C_6H_6}
\]
\( \textbf{Final answer:} \) the molecular formula is \(\mathrm{C_6H_6}\).
The empirical formula gives only the simplest ratio, so the molar mass is needed to recover the actual molecular formula.
318. The balanced equation \(2\mathrm{Mg}+\mathrm{O_2}\rightarrow2\mathrm{MgO}\) shows that:
ⓐ. \(2\,mol\) of \(\mathrm{Mg}\) reacts with \(1\,mol\) of \(\mathrm{O_2}\) to form \(2\,mol\) of \(\mathrm{MgO}\)
ⓑ. \(1\,mol\) of \(\mathrm{Mg}\) reacts with \(2\,mol\) of \(\mathrm{O_2}\) to form \(1\,mol\) of \(\mathrm{MgO}\)
ⓒ. \(2\,g\) of \(\mathrm{Mg}\) reacts with \(1\,g\) of \(\mathrm{O_2}\) to form \(2\,g\) of \(\mathrm{MgO}\)
ⓓ. \(\mathrm{Mg}\) and \(\mathrm{O_2}\) combine in any mole ratio to form \(\mathrm{MgO}\)
Correct Answer: \(2\,mol\) of \(\mathrm{Mg}\) reacts with \(1\,mol\) of \(\mathrm{O_2}\) to form \(2\,mol\) of \(\mathrm{MgO}\)
Explanation: Coefficients in a balanced chemical equation give mole ratios between reactants and products. In \(2\mathrm{Mg}+\mathrm{O_2}\rightarrow2\mathrm{MgO}\), the coefficient ratio is \(2:1:2\). This means \(2\,mol\) of magnesium reacts with \(1\,mol\) of oxygen molecules to form \(2\,mol\) of magnesium oxide. The coefficients are not gram masses. Mass relationships require conversion through molar masses, while mole ratios come directly from the balanced equation.
319. In the reaction \(\mathrm{N_2}+3\mathrm{H_2}\rightarrow2\mathrm{NH_3}\), how many moles of \(\mathrm{NH_3}\) are formed from \(4.0\,mol\) of \(\mathrm{N_2}\) when hydrogen is in excess?
ⓐ. \(2.0\,mol\)
ⓑ. \(4.0\,mol\)
ⓒ. \(8.0\,mol\)
ⓓ. \(12.0\,mol\)
Correct Answer: \(8.0\,mol\)
Explanation: \( \textbf{Balanced equation:} \) \(\mathrm{N_2}+3\mathrm{H_2}\rightarrow2\mathrm{NH_3}\).
\( \textbf{Given amount:} \) \(4.0\,mol\) of \(\mathrm{N_2}\).
\( \textbf{Condition:} \) hydrogen is in excess, so \(\mathrm{N_2}\) controls product formation.
\( \textbf{Mole ratio:} \)
\[
1\,mol\ \mathrm{N_2}\rightarrow2\,mol\ \mathrm{NH_3}
\]
\( \textbf{Apply the ratio:} \)
\[
4.0\,mol\ \mathrm{N_2}\times\frac{2\,mol\ \mathrm{NH_3}}{1\,mol\ \mathrm{N_2}}
\]
\( \textbf{Calculation:} \)
\[
n_{\mathrm{NH_3}}=8.0\,mol
\]
\( \textbf{Final answer:} \) \(8.0\,mol\) of \(\mathrm{NH_3}\) is formed.
The coefficient \(2\) for \(\mathrm{NH_3}\) must be compared with the coefficient \(1\) for \(\mathrm{N_2}\).
320. For the reaction \(2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O}\), the amount of \(\mathrm{O_2}\) needed to react exactly with \(5.0\,mol\) of \(\mathrm{H_2}\) is:
ⓐ. \(5.0\,mol\)
ⓑ. \(7.5\,mol\)
ⓒ. \(10.0\,mol\)
ⓓ. \(2.5\,mol\)
Correct Answer: \(2.5\,mol\)
Explanation: \( \textbf{Balanced equation:} \) \(2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O}\).
\( \textbf{Given amount:} \) \(5.0\,mol\) of \(\mathrm{H_2}\).
\( \textbf{Required quantity:} \) moles of \(\mathrm{O_2}\).
\( \textbf{Mole ratio from equation:} \)
\[
2\,mol\ \mathrm{H_2}:1\,mol\ \mathrm{O_2}
\]
\( \textbf{Set up conversion:} \)
\[
5.0\,mol\ \mathrm{H_2}\times\frac{1\,mol\ \mathrm{O_2}}{2\,mol\ \mathrm{H_2}}
\]
\( \textbf{Calculation:} \)
\[
n_{\mathrm{O_2}}=2.5\,mol
\]
\( \textbf{Final answer:} \) \(2.5\,mol\) of \(\mathrm{O_2}\) is required.
The oxygen coefficient is half the hydrogen coefficient, so oxygen moles are half the hydrogen moles for exact reaction.
