You can dissolve a penny using nitric acid, HNO₃. The reaction
Cu(s) + 4 HNO₃(aq) $\rarr$ Cu(NO₃) ₂(aq) + 2 NO₂(g) + 2 H₂O(l) is used to dissolve copper metal. If there is 1.000 g of copper to be dissolved, and instructions say to use four times as much acid as the required amount, how many mL (to the nearest mL) of 15.0 M nitric acid would be required to dissolve the copper? Hint: Don't forget the coefficients in the balanced reaction above when performing your calculations with nitric acid.

Question

You can dissolve a penny using nitric acid, HNO₃. The reaction
Cu(s) + 4 HNO₃(aq)

\rarr

Cu(NO₃) ₂(aq) + 2 NO₂(g) + 2 H₂O(l) is used to dissolve copper metal. If there is 1.000 g of copper to be dissolved, and instructions say to use four times as much acid as the required amount, how many mL (to the nearest mL) of 15.0 M nitric acid would be required to dissolve the copper? Hint: Don't forget the coefficients in the balanced reaction above when performing your calculations with nitric acid.

Quizplus · Accepted Answer

The balanced reaction shows that 1 mole of Cu reacts with 4 moles of HNO3. The molar mass of Cu is 63.55 g/mol, so 1.000 g of Cu is 1.000 g / 63.55 g/mol = 0.01575 mol of Cu. To dissolve this amount of Cu, we need 4 times as much HNO3, or 4 x 0.01575 mol = 0.063 mol of HNO3. The molarity of the HNO3 solution is 15.0 M, so we need 0.063 mol / 15.0 M = 0.0042 L = 4.2 mL of HNO3. To the nearest mL, this is 5 mL.

You Can Dissolve a Penny Using Nitric Acid, HNO3 →\rarr→ Cu(NO3)2(aq)+ 2 NO2(g)+ 2 H2O(l)is Used to Dissolve Copper Metal

You Can Dissolve a Penny Using Nitric Acid, HNO₃ $\rarr$ Cu(NO₃)₂(aq)+ 2 NO₂(g)+ 2 H₂O(l)is Used to Dissolve Copper Metal