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when powdered zinc is heated with sulfur, a violent reaction occurs, an…

Question

when powdered zinc is heated with sulfur, a violent reaction occurs, and zinc sulfide forms:
\\ \text{zn}(s) + \text{s}_8(s) \
ightarrow \text{zns}(s) \quad \text{unbalanced} \\
some of the reactants also combine with oxygen in air to form zinc oxide and sulfur dioxide. when \\(83.2\text{ g}\\) of \\(\text{zn}\\) reacts with \\(52.4\text{ g}\\) of \\(\text{s}_8\\), \\(109.2\text{ g}\\) of \\(\text{zns}\\) forms.

part 1 of 3
what is the percent yield of \\(\text{zns}\\)? be sure your answer has the correct number of significant figures.

part 2 of 3
if all the remaining reactants combine with oxygen, how many grams of \\(\text{zno}\\) form? be sure your answer has the correct number of significant figures.

Explanation:

Balance the main chemical equation

We balance the reaction of zinc with sulfur:

$$ 8\text{Zn}(s) + \text{S}_8(s) ightarrow 8\text{ZnS}(s) $$

Using the Chemical Equation Balancing knowledge point, the stoichiometric ratio is \(8\text{ moles of Zn} : 1\text{ mole of S}_8 : 8\text{ moles of ZnS}\).

Determine the limiting reactant

Using the Limiting Reactant Stoichiometry knowledge point, we calculate the initial moles of each reactant:

  • Molar mass of \(\text{Zn} = 65.38\text{ g/mol}\)
  • Molar mass of \(\text{S}_8 = 8 \times 32.065 = 256.52\text{ g/mol}\)
$$ n_{\text{Zn, initial}} = \frac{83.2\text{ g}}{65.38\text{ g/mol}} \approx 1.2726\text{ mol} $$
$$ n_{\text{S}_8\text{, initial}} = \frac{52.4\text{ g}}{256.52\text{ g/mol}} \approx 0.20427\text{ mol} $$

We compare the mole ratios to find the limiting reactant:

  • Moles of \(\text{Zn}\) needed to react completely with \(0.20427\text{ mol of S}_8\):
$$ 0.20427\text{ mol S}_8 \times \frac{8\text{ mol Zn}}{1\text{ mol S}_8} = 1.6342\text{ mol Zn} $$

Since we only have \(1.2726\text{ mol of Zn}\), \(\text{Zn}\) is the limiting reactant.

Calculate the percent yield of ZnS

Using the Percent Yield Calculations knowledge point, we find the theoretical yield of \(\text{ZnS}\) based on the limiting reactant \(\text{Zn}\):

  • Molar mass of \(\text{ZnS} = 65.38 + 32.065 = 97.445\text{ g/mol}\)
  • Theoretical moles of \(\text{ZnS}\) formed:
$$ n_{\text{ZnS, theoretical}} = n_{\text{Zn, initial}} = 1.2726\text{ mol} $$
  • Theoretical mass of \(\text{ZnS}\):
$$ m_{\text{ZnS, theoretical}} = 1.2726\text{ mol} \times 97.445\text{ g/mol} \approx 124.01\text{ g} $$
  • Percent yield:
$$ \text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% = \frac{109.2\text{ g}}{124.01\text{ g}} \times 100\% \approx 88.057\% $$

Rounding to three significant figures gives \(88.1\%\).

Calculate the remaining reactants

Using the Stoichiometry knowledge point, we determine how much of each reactant was consumed to form the actual \(109.2\text{ g of ZnS}\):

  • Actual moles of \(\text{ZnS}\) formed:
$$ n_{\text{ZnS, actual}} = \frac{109.2\text{ g}}{97.445\text{ g/mol}} \approx 1.1206\text{ mol} $$
  • Moles of \(\text{Zn}\) consumed:
$$ n_{\text{Zn, consumed}} = n_{\text{ZnS, actual}} = 1.1206\text{ mol} $$
  • Remaining moles of \(\text{Zn}\):
$$ n_{\text{Zn, remaining}} = 1.2726\text{ mol} - 1.1206\text{ mol} = 0.1520\text{ mol} $$
  • Moles of \(\text{S}_8\) consumed:
$$ n_{\text{S}_8\text{, consumed}} = \frac{1.1206\text{ mol}}{8} \approx 0.14008\text{ mol} $$
  • Remaining moles of \(\text{S}_8\):
$$ n_{\text{S}_8\text{, remaining}} = 0.20427\text{ mol} - 0.14008\text{ mol} = 0.06419\text{ mol} $$

Calculate the mass of ZnO formed

The remaining zinc reacts with oxygen to form zinc oxide (\(\text{ZnO}\)):

$$ 2\text{Zn} + \text{O}_2 ightarrow 2\text{ZnO} $$
  • The mole ratio of \(\text{Zn}\) to \(\text{ZnO}\) is \(1:1\).
  • Moles of \(\text{ZnO}\) formed:
$$ n_{\text{ZnO}} = n_{\text{Zn, remaining}} = 0.1520\text{ mol} $$
  • Molar mass of \(\text…

Answer:

Question 1

\(88.1\%\)

Question 2

\(12.4\text{ g}\)