Question

Aluminum reacts with excess copper(II) sulfate according to the unbalanced reaction

Al(s) + CuSO4(aq) −→
Al2(SO4)3(aq) + Cu(s)
If 2.98 g of Al react and the percent yield of
Cu is 46.4%, what mass of Cu is produced?
Answer in units of g.

Answer 1

Answer: The mass of Cu produced is 4.88 g

Step-by-step 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}=(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 =
`(3)/(2)* 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* 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}}* 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}* 100\\\\\text{Actual value of Cu}=(46.4* 10.516)/(100)\\\\\text{Actual value of Cu}=4.88g`

Hence, the mass of Cu produced is 4.88 g