Question

100mL of 5mol L-1 nitric acid was added to excess copper. What mass of copper reacted? Cu +4HNO3 -> 2NO2 + Cu(NO3)2

Answer 1

The mass of `"Cu"` was d) 7.9 g.

Explanation:

Step 1. Start with the balanced equation.

`"Cu" + "4HNO"_3 → "2NO"_2 + "Cu"("NO"_3)_2`

Step 2. Calculate the moles of `"HNO"_3`.

`"Moles of HNO"_3 = 0.100 color(red)(cancel(color(black)("L HNO"_3))) × ("5 mol HNO"_3)/(1 color(red)(cancel(color(black)("L HNO"_3)))) = "0.50 mol HNO"_3`

Step 3. Calculate the moles of `"Cu"`

`"Moles of Cu" = 0.50 color(red)(cancel(color(black)("mol HNO"_3))) × "1 mol Cu"/(4color(red)(cancel(color(black)("mol HNO"_3)))) = "0.12 mol Cu"`

4. Calculate the mass of `"Cu"`.

`"Mass of Cu" = 0.12 color(red)(cancel(color(black)("mol Cu"))) × "63.55 g Cu"/(1 color(red)(cancel(color(black)("mol Cu")))) = "7.9 g Cu"`

Note: The answer can have only 1 significant figure, because that is all you gave for the concentration of the nitric acid.

However, I calculated the answer to two significant figures for you.