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How many mL of 16 M HNO₃ is required to react with 0.5 g of copper metal?

a) 0.019
b) 0.029
c) 0.039
d) 0.049

1 Answer

4 votes

Final answer:

To react with 0.5 g of copper metal, 0.01884 mL of 16 M HNO₃ is required.

Step-by-step explanation:

To find the volume of 16 M HNO₃ required to react with 0.5 g of copper metal, we need to use the balanced chemical equation:

3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O + 2NO

From the equation, we can see that 3 moles of copper react with 8 moles of HNO₃.

First, calculate the moles of copper using its molar mass (63.55 g/mol):

Moles of copper = Mass of copper / Molar mass of copper = 0.5 g / 63.55 g/mol = 0.007876 mol

Since the ratio of moles of copper to moles of HNO₃ is 3:8, we can calculate the moles of HNO₃:

Moles of HNO₃ = Moles of copper x (8 moles HNO₃ / 3 moles copper) = 0.007876 mol x (8 mol HNO₃ / 3 mol copper) = 0.0210 mol HNO₃

To find the volume of HNO₃, we can use the formula:

Volume (L) = Moles / Molarity

Volume of 16 M HNO₃ = 0.0210 mol / 16 M = 0.001884 L = 0.01884 mL

User Robertas Uldukis
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