Question

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Answer 1

Answer: The mass of Cu produced is 4.88 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ……(1)

Given mass of aluminum = 2.98 g

Molar mass of aluminum = 27 g/mol

Plugging values in equation 1:

\text{Moles of aluminum}=\frac{2.98g}{27g/mol}=0.1104 mol

The given chemical equation follows:

2Al(s)+3CuSO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3Cu(s)

By the stoichiometry of the reaction:

If 2 moles of aluminum produces 3 moles of Cu

So, 0.1104 moles aluminium will produce = \frac{3}{2}\times 0.1104=0.1656mol of Cu

Molar mass of Cu = 63.5 g/mol

Plugging values in equation 1:

\text{Mass of Cu}=(0.1656mol\times 63.5g/mol)=10.516g

The percent yield of a reaction is calculated by using an equation:

\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100 ……(2)

Given values:

% yield of product = 46.4 %

Theoretical value of the product = 10.516 g

Plugging values in equation 2, we get:

46.4=\frac{\text{Actual value of Cu}}{10.516g}\times 100\\\\\text{Actual value of Cu}=\frac{46.4\times 10.516}{100}\\\\\text{Actual value of Cu}=4.88g

Hence, the mass of Cu produced is 4.88 g

Answer 2

2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu

mole Al = 2.57 : 27 g/mole = 0.095

mole Cu = 3/2 x mole Al = 3/2 x 0.095 = 0.143

mass Cu (theoretical) = 0.143 x molar mass Cu = 0.143 x 63.5 g/mole = 9.08 g

mass Cu produced (actual, 56.9% yield) = 56.9% x 9.08 g = 5.17 g