Mole ratios: Chemistry Regents Practice

Given the balanced equation $2H_2 + O_2 \rightarrow 2H_2O$, how many moles of $O_2$ are required to react completely with $4.0$ moles of $H_2$?

• (A) $1.0$ mol $O_2$
• (B) $4.0$ mol $O_2$
• (C) $8.0$ mol $O_2$
• (D) $2.0$ mol $O_2$ ✓

The coefficients of a balanced equation give the mole ratio; here $H_2:O_2 = 2:1$, so half as many moles of $O_2$ are needed.

Propane burns according to $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$. At STP, what volume of $CO_2$ is produced when $2.0$ liters of $C_3H_8$ react completely?

• (A) $0.67$ L $CO_2$
• (B) $2.0$ L $CO_2$
• (C) $10.$ L $CO_2$
• (D) $6.0$ L $CO_2$ ✓

For gases at the same temperature and pressure, volume ratios equal mole (coefficient) ratios, so $1$ volume of $C_3H_8$ yields $3$ volumes of $CO_2$.

Given the balanced equation: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$. If 5.00 moles of $N_2$ react completely with excess $H_2$, determine the number of moles of $NH_3$ produced.

Coefficients in a balanced chemical equation give the mole ratio between reactants and products, which is used to convert moles of a known substance to moles of a desired substance.

Given the balanced equation: $2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$. If 5.0 moles of $H_2$ react completely with excess oxygen, determine the number of grams of $H_2O$ produced. The gram-formula mass of $H_2O$ is 18.0 g/mol.

The coefficients of a balanced equation give the mole ratio between reactants and products; here the 1:1 ratio of H2 to H2O means 5.0 moles of H2 produce 5.0 moles of H2O, which is then converted to grams using the gram-formula mass.

Given the balanced equation: $2Al + 3Cl_2 \rightarrow 2AlCl_3$. Determine the number of moles of $Cl_2$ needed to react completely with 5.0 moles of $Al$.

Mole ratios from the coefficients of a balanced equation allow conversion between moles of any two substances in the reaction; here the Al-to-Cl2 ratio of 2:3 is applied to 5.0 mol Al.

Given the balanced equation: $2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$. What is the total number of moles of $H_2O$ produced when 6.0 moles of $H_2$ react completely with excess $O_2$?

• (A) 3.0 mol
• (B) 9.0 mol
• (C) 12.0 mol
• (D) 6.0 mol ✓

The coefficients of H2 and H2O in the balanced equation are equal (2:2), so the moles of H2O produced equal the moles of H2 that reacted.

Given the balanced equation $2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$, how many moles of water form from $4.0$ moles of $\mathrm{H}_2$ (excess oxygen)?

• (A) $1.0\ \mathrm{mol}$
• (B) $2.0\ \mathrm{mol}$
• (C) $4.0\ \mathrm{mol}$ ✓
• (D) $8.0\ \mathrm{mol}$

Use the balanced coefficients as a mole ratio: multiply the known amount by (moles wanted ÷ moles known).

For $\mathrm{N}_2 + 3\mathrm{H}_2 \rightarrow 2\mathrm{NH}_3$, how many moles of $\mathrm{NH}_3$ form from $6.0$ mol of $\mathrm{H}_2$ (excess $\mathrm{N}_2$)?

• (A) $2.0\ \mathrm{mol}$
• (B) $6.0\ \mathrm{mol}$
• (C) $9.0\ \mathrm{mol}$
• (D) $4.0\ \mathrm{mol}$ ✓

Multiply the known moles by the coefficient ratio taken straight from the balanced equation.

In $2\mathrm{Na} + \mathrm{Cl}_2 \rightarrow 2\mathrm{NaCl}$, how many moles of $\mathrm{NaCl}$ form from $3.0$ mol of $\mathrm{Na}$ (excess $\mathrm{Cl}_2$)?

• (A) $2.0\ \mathrm{mol}$
• (B) $3.0\ \mathrm{mol}$ ✓
• (C) $6.0\ \mathrm{mol}$
• (D) $1.5\ \mathrm{mol}$

Equal coefficients give a 1:1 mole ratio, so the amount carries straight through.

Given $\;2H_2+O_2\rightarrow 2H_2O$, how many moles of water form from $4.0$ mol of $H_2$ (with excess $O_2$)?

• (A) $4.0$ mol ✓
• (B) $8.0$ mol
• (C) $2.0$ mol
• (D) $1.0$ mol

Use the coefficient ratio: 2 mol H₂ : 2 mol H₂O is 1:1, so 4.0 mol H₂ yields 4.0 mol water.

In the reaction $N_2 + 3H_2 \rightarrow 2NH_3$, how many moles of $NH_3$ are produced when $6.0$ moles of $H_2$ react completely?

• (A) $9.0$ mol $NH_3$
• (B) $6.0$ mol $NH_3$
• (C) $2.0$ mol $NH_3$
• (D) $4.0$ mol $NH_3$ ✓

Balanced-equation coefficients set the mole ratio; multiply the given moles of $H_2$ by $\frac{2\ NH_3}{3\ H_2}$ to find moles of product.

Magnesium burns in oxygen: $2Mg + O_2 \rightarrow 2MgO$. What mass of $MgO$ is produced when $3.0$ moles of $Mg$ react completely?

• (A) $242$ g $MgO$
• (B) $73$ g $MgO$
• (C) $121$ g $MgO$ ✓
• (D) $60.$ g $MgO$

Use the $1:1$ mole ratio to find moles of $MgO$, then convert to grams with the molar mass of $MgO$ ($40.31$ g/mol), not that of $Mg$ alone.

Given the balanced equation $2Al + 3Cl_2 \rightarrow 2AlCl_3$, how many moles of $Cl_2$ are required to react completely with 4.0 moles of Al?

• (A) 4.0 mol
• (B) 8.0 mol
• (C) 2.7 mol
• (D) 6.0 mol ✓

Using the mole ratio from the balanced equation (3 mol Cl2 for every 2 mol Al), multiplying 4.0 mol Al by 3/2 gives 6.0 mol Cl2 needed. A common error is inverting the ratio (giving 2.7) or using a 1:1 ratio (giving 4.0).

Given the balanced equation: $2Al(s) + 3CuSO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 3Cu(s)$. How many moles of $Cu(s)$ are produced when 4.0 moles of $Al(s)$ react completely with excess $CuSO_4(aq)$?

• (A) 4.0 mol
• (B) 3.0 mol
• (C) 8.0 mol
• (D) 6.0 mol ✓

Using the mole ratio from the balanced equation (2 mol Al produce 3 mol Cu), 4.0 mol Al yields 6.0 mol Cu; assuming a 1:1 ratio or misapplying the coefficients gives the incorrect distractor values.

Given the balanced equation: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$. What is the maximum number of moles of $NH_3$ that can be produced from 9.0 moles of $H_2$, assuming excess $N_2$ is available?

• (A) 9.0 mol
• (B) 3.0 mol
• (C) 6.0 mol ✓
• (D) 18 mol

Using the mole ratio from the balanced equation (3 mol H2 to 2 mol NH3), 9.0 mol H2 produces 6.0 mol NH3. A common error is using the ratio inverted (multiplying by 3/2) or assuming a 1:1 mole ratio.

Given the balanced equation: $2H_2 + O_2 \rightarrow 2H_2O$. How many moles of $O_2$ are required to react completely with 6.0 moles of $H_2$?

• (A) 12.0 mol
• (B) 6.0 mol
• (C) 3.0 mol ✓
• (D) 1.5 mol

The coefficients of a balanced equation give the mole ratio between reactants; since the ratio of H2 to O2 is 2:1, 6.0 moles of H2 requires half that amount, 3.0 moles, of O2. Answers of 6.0 or 12.0 mol come from misapplying the ratio (using it as 1:1 or 1:2 instead of 2:1).

Given the balanced equation: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$. Determine the number of moles of $H_2$ gas needed to react completely with 4.0 moles of $N_2$ gas.

The coefficients of a balanced equation give the mole ratio between reactants. Multiplying the given moles of N2 by the mole ratio of H2 to N2 (3:1) gives the moles of H2 required.

Aluminum reacts with sulfuric acid according to the balanced equation: $2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2$. If 4.0 moles of Al react completely, determine the number of moles of $H_2$ gas produced.

The coefficients in a balanced equation give the mole ratio between reactants and products; multiplying the given moles of Al by the ratio of H2 to Al (3:2) yields the moles of H2 produced.

Nitrogen gas reacts with hydrogen gas to produce ammonia according to the balanced equation: $N_2 + 3H_2 \rightarrow 2NH_3$. If 9.0 moles of $H_2$ react completely, determine the number of moles of $NH_3$ produced.

Stoichiometry problems use the coefficients of a balanced equation as a mole ratio (conversion factor) to convert from moles of a given substance to moles of a desired substance.

Given the balanced equation $2H_2 + O_2 \rightarrow 2H_2O$, how many moles of $O_2$ are required to react completely with 6.0 moles of $H_2$?

• (A) 1.5 mol
• (B) 3.0 mol ✓
• (C) 12 mol
• (D) 6.0 mol

Using the mole ratio from the balanced equation (1 mol O2 for every 2 mol H2), 6.0 mol H2 requires 3.0 mol O2.

Given the balanced equation: $2H_2 + O_2 \rightarrow 2H_2O$. If 4.0 moles of $H_2$ react completely, how many moles of $O_2$ are consumed?

• (A) 8.0 mol
• (B) 2.0 mol ✓
• (C) 1.0 mol
• (D) 4.0 mol

The coefficients of a balanced equation give the mole ratio between reactants. Since 2 mol H2 reacts with 1 mol O2, 4.0 mol H2 requires half that amount, 2.0 mol O2. A common mistake is using the coefficients backward or assuming a 1:1 ratio.

Given the balanced equation: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$. What is the total number of moles of $NH_3(g)$ produced when 6.0 moles of $H_2(g)$ are completely consumed?

• (A) 4.0 moles ✓
• (B) 6.0 moles
• (C) 3.0 moles
• (D) 2.0 moles

The coefficients of a balanced equation give the mole ratio between substances. Using the 3:2 ratio of H2 to NH3, 6.0 moles of H2 produces 4.0 moles of NH3.

Given the balanced equation: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$. What is the number of moles of $H_2(g)$ needed to react completely with 5.0 moles of $N_2(g)$?

• (A) 15 mol ✓
• (B) 5.0 mol
• (C) 10. mol
• (D) 20. mol

The coefficients of a balanced equation give the mole ratio between reactants. Since 3 mol of H2 react with every 1 mol of N2, 5.0 mol N2 requires 15 mol H2. A common error is forgetting to multiply by the ratio (choosing 5.0 mol) or doubling incorrectly (choosing 10. mol).

Given the balanced equation $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$, if 5.00 moles of $N_2$ react completely with excess hydrogen gas, calculate the number of moles of $NH_3$ produced.

Mole ratios come directly from the coefficients of a balanced chemical equation. Multiplying the given moles of N2 by the ratio of NH3 to N2 (2:1) converts moles of reactant to moles of product.