Question

Given the following balanced equation:

3Cu(s) + 8HNO3(aq) = 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
Determine the mass of copper (II) nitrate that would be formed from the complete reaction
of 35.5g of copper with an excess of nitric acid.

Answer 1

Answer: The mass of copper (II) nitrate produced is 105.04 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 copper = 35.5 g

Molar mass of copper = 63.5 g/mol

Plugging values in equation 1:

`\text{Moles of copper}=(35.5g)/(63.5g/mol)=0.560 mol`

The given chemical equation follows:

`3Cu(s)+8HNO_3(aq)\rightarrow 3Cu(NO_3)_2(aq)+2NO(g)+4H_2O(l)`

By the stoichiometry of the reaction:

If 3 moles of copper produces 3 moles of copper (II) nitrate

So, 0.560 moles of copper will produce =
`(3)/(3)* 0.560=0.560mol` of copper (II) nitrate

Molar mass of copper (II) nitrate = 187.56 g/mol

Plugging values in equation 1:

`\text{Mass of copper (II) nitrate}=(0.560mol* 187.56g/mol)=105.04g`

Hence, the mass of copper (II) nitrate produced is 105.04 g.