Questions: Application Practice Problems: Limiting Reactant Stoichiometry Instructions: Complete the following conversions, showing all of the steps in your work. Copper (I) sulfide is produced from the synthesis of copper and sulfur 2 Cu + S → CuS 2. What is the limiting reactant when 80.0 g Cu reacts with 25.0 g S 80.0 Cu × 1 mol Cu / 63.55 g = 1.25 mol × 1 mol S / 2 mol Cu = 0.830 mol S 25.0 g S × 1 mol S / 32.07 g = 0.779 mol × 2 mol Cu / 1 mol S = 1.558 mol Cu Cu is limiting b. What is the maximum number of grams of Cu2S that can be formed when starting with the quantities of each reactant given above? c. How many grams of the excess reactant will remain after the limiting reactant is consumed?

Application Practice Problems: Limiting Reactant Stoichiometry Instructions: Complete the following conversions, showing all of the steps in your work. Copper (I) sulfide is produced from the synthesis of copper and sulfur 2 Cu + S → CuS 2. What is the limiting reactant when 80.0 g Cu reacts with 25.0 g S 80.0 Cu × 1 mol Cu / 63.55 g = 1.25 mol × 1 mol S / 2 mol Cu = 0.830 mol S 25.0 g S × 1 mol S / 32.07 g = 0.779 mol × 2 mol Cu / 1 mol S = 1.558 mol Cu Cu is limiting b. What is the maximum number of grams of Cu2S that can be formed when starting with the quantities of each reactant given above? c. How many grams of the excess reactant will remain after the limiting reactant is consumed?
Transcript text: Application Practice Problems: Limiting Reactant Stoichiometry Instructions: Complete the following conversions, showing all of the steps in your work. Copper (I) sulfide is produced from the synthesis of copper and sulfur \[ 2 \mathrm{Cu}+\mathrm{s} \rightarrow \mathrm{Cus} \] \[ \begin{array}{l} \text { 2. What is the limiting reactant when } 80.0 \mathrm{~g} \text { Cu reacts with } 25.0 \mathrm{~g} S \\ 80.0, \mathrm{Cu} \times \frac{1 \mathrm{~mol} \mathrm{Cu}}{63.55 \mathrm{~g}}=1.25 \mathrm{mol} \times \frac{1 \mathrm{mol} S}{2 \mathrm{mol Cu}}=0.830 \mathrm{mol S} \\ 25.0 \mathrm{~g} S \times \frac{1 \mathrm{~mol} S}{32.07 \mathrm{~g}}=0.779 \mathrm{mol} \times \frac{2 \mathrm{mol Cu}}{1 \mathrm{mol S}}=1.558 \mathrm{mol Cu} \end{array} \] Cu is limiting b. What is the maximum number of grams of $\mathrm{Cu}_{2}S$ that can be formed when starting with the quantities of each reactant given above? c. How many grams of the excess reactant will remain after the limiting reactant is consumed?
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Solution

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Determine the limiting reactant when 80.0 g Cu reacts with 25.0 g S.

Convert grams of Cu to moles of Cu.

\[ 80.0 \, \text{g Cu} \times \frac{1 \, \text{mol Cu}}{63.55 \, \text{g}} = 1.26 \, \text{mol Cu} \]

Convert moles of Cu to moles of CuS.

\[ 1.26 \, \text{mol Cu} \times \frac{1 \, \text{mol CuS}}{2 \, \text{mol Cu}} = 0.630 \, \text{mol CuS} \]

Convert grams of S to moles of S.

\[ 25.0 \, \text{g S} \times \frac{1 \, \text{mol S}}{32.07 \, \text{g}} = 0.779 \, \text{mol S} \]

Convert moles of S to moles of CuS.

\[ 0.779 \, \text{mol S} \times \frac{1 \, \text{mol CuS}}{1 \, \text{mol S}} = 0.779 \, \text{mol CuS} \]

Determine the limiting reactant.

Since 0.630 mol CuS (from Cu) is less than 0.779 mol CuS (from S), Cu is the limiting reactant.

\(\boxed{\text{Cu is limiting}}\)

Calculate the maximum number of grams of CuS that can be formed.

Determine the moles of CuS formed from the limiting reactant.

The moles of CuS formed is 0.630 mol (from the limiting reactant Cu).

Convert moles of CuS to grams of CuS.

\[ 0.630 \, \text{mol CuS} \times \frac{95.61 \, \text{g}}{1 \, \text{mol CuS}} = 60.2 \, \text{g CuS} \]

\(\boxed{60.2 \, \text{g CuS}}\)

Calculate the grams of the excess reactant (S) that will remain.

Determine the moles of S that react with the limiting reactant.

\[ 1.26 \, \text{mol Cu} \times \frac{1 \, \text{mol S}}{2 \, \text{mol Cu}} = 0.630 \, \text{mol S} \]

Calculate the initial moles of S.

The initial moles of S is 0.779 mol.

Calculate the moles of S remaining.

\[ 0.779 \, \text{mol S} - 0.630 \, \text{mol S} = 0.149 \, \text{mol S} \]

Convert the remaining moles of S to grams.

\[ 0.149 \, \text{mol S} \times \frac{32.07 \, \text{g}}{1 \, \text{mol S}} = 4.78 \, \text{g S} \]

\(\boxed{4.78 \, \text{g S}}\)

\(\boxed{\text{Cu is limiting}}\)

\(\boxed{60.2 \, \text{g CuS}}\)

\(\boxed{4.78 \, \text{g S}}\)

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