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Q
QuestionChemistry

``` molar mass AI = 26.98 g/mol molar mass HCl = 36.45 g/mol molar mass AlCl^3 = 133.33 g/mol molar mass H^2 = 2.00 g/mol ``` If 4.5 grams of hydrogen chloride completely react according to the provided equation and 3.2 grams of aluminum chloride are collected, what is the percent yield?
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Answer

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Step 1:
I'll solve this step by step, using precise LaTeX formatting:

Step 2:
: Write the balanced chemical equation

2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_{3} + 3 \text{H}_{2}

Step 3:
: Calculate moles of HCl used

n_{\text{HCl}} = \frac{4.5 \text{ g}}{36.45 \text{ g/mol}} = 0.1235 \text{ mol}

Step 4:
: Determine limiting reactant stoichiometry

\frac{0.1235 \text{ mol HCl} \times (2 \text{ mol AlCl}_{3})}{(6 \text{ mol HCl})} = 0.0412 \text{ mol AlCl}_{3} \text{ theoretical}
From the balanced equation, 6 mol HCl produces 2 mol AlCl₃

Step 5:
: Calculate theoretical mass of AlCl₃

m_{\text{theoretical}} = 0.0412 \text{ mol} \times 133.33 \text{ g/mol} = 5.50 \text{ g}

Step 6:
: Calculate percent yield

\text{Percent Yield} = \frac{3.2 \text{ g (actual)}}{5.50 \text{ g (theoretical)}} \times 100\% = 58.2\%

Final Answer

The percent yield is 58.2%.