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

Answer:

[tex]m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2[/tex]

Explanation:

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

[tex]2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2[/tex]

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:

[tex]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[/tex]

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:

[tex]4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2[/tex]

Best regards!

Amount of copper (II) nitrate would be produced is 2.25 gram

Given that;

Amount of copper metal = 0.80 g

Find:

Amount of copper (II) nitrate would be produced

Computation:

Cu + 4HNO₃ → Cu(NO₃) + 2NO₂ + 2H₂O

Number of moles = Given mass / Molecular mass

Number of moles = 0.80 / 63.5

Number of moles = 0.012 mol

So,

n = Given mass / Molecular mass

0.012 = Given mass / 63.5 + (63.5 + 48)

Given mass = 2.25 gram

Amount of copper (II) nitrate would be produced = 2.25 gram

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