Question

missed this? watch kcv: limiting reactant, theoretical yield, and percent yield. iwe: finding limiting reactant and theoretical yield. read section 8.6. you can click on the review link to access the section in your etext. consider the reaction between nis₂ and o₂: 2 nis₂ (s) + 5 o₂ (g) → 2 nio (s) + 4 so₂ (g) when 11.2 g of nis₂ are allowed to react with 5.43 g of o₂, 4.44 g of nio are obtained. part a determine the limiting reactant for the reaction. express your answer as a chemical formula. part b determine the theoretical yield of nio for the reaction.

Explanation:

Step1: Calculate moles of \( \text{NiS}_2 \)

Molar mass of \( \text{NiS}_2 \): \( \text{Ni} = 58.69 \, \text{g/mol} \), \( \text{S} = 32.07 \, \text{g/mol} \), so \( M = 58.69 + 2\times32.07 = 122.83 \, \text{g/mol} \).
Moles of \( \text{NiS}_2 = \frac{11.2 \, \text{g}}{122.83 \, \text{g/mol}} \approx 0.0912 \, \text{mol} \).

Step2: Calculate moles of \( \text{O}_2 \)

Molar mass of \( \text{O}_2 = 32.00 \, \text{g/mol} \).
Moles of \( \text{O}_2 = \frac{5.43 \, \text{g}}{32.00 \, \text{g/mol}} \approx 0.170 \, \text{mol} \).

Step3: Determine mole ratio from reaction

Reaction: \( 2\text{NiS}_2 + 5\text{O}_2 ightarrow 2\text{NiO} + 4\text{SO}_2 \).
Mole ratio \( \text{NiS}_2 : \text{O}_2 = 2:5 \).

Step4: Find required moles of \( \text{O}_2 \) for \( \text{NiS}_2 \)

For \( 0.0912 \, \text{mol} \, \text{NiS}_2 \), required \( \text{O}_2 = 0.0912 \times \frac{5}{2} = 0.228 \, \text{mol} \).

Step5: Compare available and required \( \text{O}_2 \)

Available \( \text{O}_2 = 0.170 \, \text{mol} < 0.228 \, \text{mol} \). So \( \text{O}_2 \) is limiting? Wait, no—wait, check \( \text{NiS}_2 \) required for \( \text{O}_2 \).
For \( 0.170 \, \text{mol} \, \text{O}_2 \), required \( \text{NiS}_2 = 0.170 \times \frac{2}{5} = 0.068 \, \text{mol} \).
Available \( \text{NiS}_2 = 0.0912 \, \text{mol} > 0.068 \, \text{mol} \). Thus, \( \text{O}_2 \) is limiting? Wait, no—wait, the limiting reactant is the one that runs out first. Wait, let's re-express:

From \( \text{NiS}_2 \): moles of \( \text{NiO} \) produced (if \( \text{NiS}_2 \) is limiting) = \( 0.0912 \, \text{mol} \, \text{NiS}_2 \times \frac{2}{2} = 0.0912 \, \text{mol} \, \text{NiO} \).

From \( \text{O}_2 \): moles of \( \text{NiO} \) produced (if \( \text{O}_2 \) is limiting) = \( 0.170 \, \text{mol} \, \text{O}_2 \times \frac{2}{5} = 0.068 \, \text{mol} \, \text{NiO} \).

Since \( 0.068 < 0.0912 \), \( \text{O}_2 \) limits the reaction? Wait, no—wait, the limiting reactant is the one that produces less product. Wait, no: the limiting reactant is the one that, when consumed, produces the least product. Wait, no, the limiting reactant is the reactant that is completely consumed, limiting the reaction. So if we use \( \text{O}_2 \) (0.170 mol), it will react with \( 0.068 \, \text{mol} \, \text{NiS}_2 \), leaving \( \text{NiS}_2 \) in excess. Thus, \( \text{O}_2 \) is the limiting reactant? Wait, but let's check the problem again. Wait, maybe I made a mistake. Wait, the reaction is \( 2\text{NiS}_2 + 5\text{O}_2 ightarrow 2\text{NiO} + 4\text{SO}_2 \). So 2 moles \( \text{NiS}_2 \) react with 5 moles \( \text{O}_2 \).

Wait, let's recalculate moles:

\( \text{NiS}_2 \): 11.2 g / 122.83 g/mol ≈ 0.0912 mol (correct).

\( \text{O}_2 \): 5.43 g / 32.00 g/mol ≈ 0.170 mol (correct).

Mole ratio \( \text{NiS}_2 : \text{O}_2 \) in reaction: 2:5 = 0.4:1.

Actual mole ratio: 0.0912 / 0.170 ≈ 0.536:1.

Since 0.536 > 0.4 (required ratio), \( \text{NiS}_2 \) is in excess, so \( \text{O}_2 \) is limiting. Wait, no—wait, the required ratio for \( \text{NiS}_2 \) to \( \text{O}_2 \) is 2/5 = 0.4. If actual ratio (NiS2/O2) is greater than 0.4, that means there's more NiS2 than needed for O2, so O2 is limiting. Yes. So limiting reactant is \( \text{O}_2 \)? Wait, but let's check the theoretical yield using limiting reactant.

Theoretical yield of \( \text{NiO} \): from \( \text{O}_2 \), moles of \( \text{NiO} = 0.170 \, \text{mol} \, \text{O}_2 \times \frac{2}{5} = 0.068 \, \text{mol} \).

Molar mass of \( \text{NiO} = 58.69 + 16.00 = 74.69 \, \text{g/mol} \).

Theoretical yield = \( 0.068 \, \text{mol} \times 74.69 \, \text{g/mol} \approx 5.08 \, \text{g} \)? Wait, but the actual yield is 4.44 g. Wait, but Part B is theoretical yield. Wait, maybe I messed up limiting reactant. Wait, let's re-express:

Wait, maybe I had it backwards. Let's calculate moles of each reactant:

\( \text{NiS}_2 \): 11.2 g / 122.83 g/mol ≈ 0.0912 mol.

\( \text{O}_2 \): 5.43 g / 32.00 g/mol ≈ 0.170 mol.

Reaction: 2 NiS2 + 5 O2 → 2 NiO + 4 SO2.

Mole ratio NiS2:O2 = 2:5 = 0.4:1.

Actual ratio NiS2:O2 = 0.0912 / 0.170 ≈ 0.536:1. Since 0.536 > 0.4, NiS2 is in excess, O2 is limiting. So O2 is the limiting reactant.

Thus, moles of NiO produced = (2/5) * moles of O2 = (2/5)*0.170 = 0.068 mol.

Theoretical yield of NiO = 0.068 mol * 74.69 g/mol ≈ 5.08 g? Wait, but the actual yield is 4.44 g. Wait, maybe I made a mistake in limiting reactant. Wait, let's check with NiS2 as limiting:

Moles of O2 needed for NiS2: 0.0912 mol NiS2 * (5/2) = 0.228 mol O2. But we only have 0.170 mol O2, which is less than 0.228 mol. So O2 is limiting, so NiS2 is in excess. So theoretical yield is based on O2.

Wait, but let's recalculate molar mass of NiS2: Ni=58.69, S=32.07*2=64.14, total=58.69+64.14=122.83 g/mol. Correct.

O2 molar mass: 32.00 g/mol. Correct.

Moles of O2: 5.43 / 32 = 0.170 mol. Correct.

Moles of NiO from O2: (2/5)*0.170 = 0.068 mol.

Theoretical yield: 0.068 mol * 74.69 g/mol = 0.068*74.69 ≈ 5.08 g. Wait, but maybe I messed up the limiting reactant. Wait, let's check the other way: if we use NiS2 as limiting, moles of NiO would be 0.0912 mol (since 2 NiS2 → 2 NiO), so mass = 0.0912 * 74.69 ≈ 6.81 g. But we only have 0.170 mol O2, which can only produce 0.068 mol NiO (5.08 g). So O2 is limiting, so theoretical yield is ~5.08 g. Wait, but maybe I made a mistake in the ratio. Wait, reaction is 2 NiS2 + 5 O2 → 2 NiO + 4 SO2. So 2 moles NiS2 react with 5 moles O2 to produce 2 moles NiO. So mole ratio NiO:O2 = 2:5. So moles NiO = (2/5)*moles O2. Correct.

Alternatively, maybe the limiting reactant is NiS2? Wait, no—because 0.0912 mol NiS2 needs 0.228 mol O2, but we have only 0.170 mol O2, so O2 is insufficient, so O2 is limiting. Therefore, theoretical yield is based on O2.

Wait, but let's check the problem again. The problem says "Determine the theoretical yield of NiO for the reaction." So we need to find which reactant is limiting, then calculate.

Wait, maybe I made a mistake in moles of NiS2. Let's recalculate: 11.2 g / 122.83 g/mol = 0.0912 mol (correct). 5.43 g O2 / 32 g/mol = 0.170 mol (correct).

Reaction stoichiometry: 2 mol NiS2 reacts with 5 mol O2 to produce 2 mol NiO.

So, the limiting reactant is the one with the smaller mole ratio (actual moles / stoichiometric coefficient).

For NiS2: 0.0912 / 2 = 0.0456.

For O2: 0.170 / 5 = 0.034.

Since 0.034 < 0.0456, O2 has the smaller ratio, so O2 is limiting.

Thus, moles of NiO = (2/5) * moles of O2 = (2/5)*0.170 = 0.068 mol.

Theoretical yield = 0.068 mol * 74.69 g/mol ≈ 5.08 g. Wait, but let's check with exact calculation:

0.170 mol O2 * (2 mol NiO / 5 mol O2) * 74.69 g/mol NiO = 0.170 * (2/5) * 74.69 = 0.068 * 74.69 = 5.07892 ≈ 5.08 g.

So the theoretical yield is approximately 5.08 g.

Answer:

The theoretical yield of \( \text{NiO} \) is approximately \( \boldsymbol{5.08 \, \text{g}} \) (or more precisely, follow the calculation steps for exact value).