Question

question 34 (1 point)
consider the reaction: 3cucl2(aq) + 2al(s) → 3cu(s) + 2alcl3(aq). what is the theoretical yield of solid copper when 3.89 g of copper(ii) chloride is allowed to react in the presence of excess of aluminum?
3.89 g
0.0289 g
8.23 g
1.84 g

Explanation:

Step1: Calculate the molar mass of \(CuCl_2\)

The molar mass of \(Cu\) is \(63.55\ g/mol\), and the molar mass of \(Cl\) is \(35.45\ g/mol\). For \(CuCl_2\), \(M = 63.55+(2\times35.45)=134.45\ g/mol\).

Step2: Find the moles of \(CuCl_2\)

Given mass of \(CuCl_2\) is \(m = 3.89\ g\). Using the formula \(n=\frac{m}{M}\), \(n=\frac{3.89\ g}{134.45\ g/mol}\approx0.0289\ mol\).

Step3: Use the stoichiometry of the reaction

From the balanced equation \(3CuCl_2(aq)+2Al(s)\to3Cu(s)+2AlCl_3(aq)\), the mole ratio of \(CuCl_2\) to \(Cu\) is \(3:3 = 1:1\). So moles of \(Cu\) produced \(n_{Cu}=n_{CuCl_2}\approx0.0289\ mol\).

Step4: Calculate the mass of \(Cu\)

The molar mass of \(Cu\) is \(M_{Cu}=63.55\ g/mol\). Using \(m = n\times M\), \(m_{Cu}=0.0289\ mol\times63.55\ g/mol\approx1.84\ g\).

Answer:

\(1.84\ g\)