Question

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we finding limiting reactant and theoretical yield: read section 8.6. you can click on
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for the reaction shown, compute the theoretical yield of the product in grams for each of
the following initial amounts of reactants.

$$ 2 mathrm { al } ( s ) + 3 mathrm { cl } _ { 2 } ( g ) ightarrow 2 mathrm { alcl } _ { 3 } ( s ) $$

part a
1.0 g al, 1.0 g cl2
express your answer using two significant figures.
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Explanation:

Step1: Calculate moles of reactants

Molar mass of \(Al = 26.98\ g/mol\), moles of \(Al=\frac{1.0\ g}{26.98\ g/mol}\approx0.037\ mol\)
Molar mass of \(Cl_2 = 70.90\ g/mol\), moles of \(Cl_2=\frac{1.0\ g}{70.90\ g/mol}\approx0.0141\ mol\)

Step2: Determine limiting reactant

From the balanced equation \(2Al(s)+3Cl_2(g)\to2AlCl_3(s)\), mole ratio of \(Al:Cl_2 = 2:3\).
For \(0.037\ mol\ Al\), required \(Cl_2\) moles \(=\frac{3}{2}\times0.037\ mol = 0.0555\ mol\). Since available \(Cl_2\) is \(0.0141\ mol\lt0.0555\ mol\), \(Cl_2\) is limiting.

Step3: Calculate moles of product

Using \(Cl_2\) as limiting reactant, mole ratio of \(Cl_2:AlCl_3 = 3:2\). Moles of \(AlCl_3=\frac{2}{3}\times0.0141\ mol\approx0.0094\ mol\)

Step4: Calculate mass of product

Molar mass of \(AlCl_3=133.34\ g/mol\), mass of \(AlCl_3 = 0.0094\ mol\times133.34\ g/mol\approx1.25\ g\)

Answer:

\(1.3\ g\)