Question

How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?

Answer 1

Answer:

$m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

$2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2$

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

$m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

$4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2$

Best regards!