This is Chemistry. The equation for the reaction of copper and nitric acid

Question

The equation for the reaction of copper and dilute nitric acid is:

3Cu + 8 HNO3 yields 3 CU(no3)2 + 2 NO + 4 H20.

What volume (in mL) of a nitric acid solution which contains 27.0% HNO3 by mass and has a density of 1.19 g/m: will be needed to react with 9.35 g of Cu?

I do not know how to integrate the % acid solution into the equation.

Answer 1

3Cu + 8HNO3 ==> 3Cu(NO3)2 + 2NO + 4H2O

Find moles of HNO3 needed to react with 9.35 g of Cu:

9.35 g Cu x 1 mol Cu/63.5 g x 8 mol HNO3/3 mol Cu = 0.393 moles HNO3 required

If this were 100% pure, calculate mass needed:

0.393 mol HNO3 x 63.01 g/mol = 24.7 g HNO3 if were 100% pure

Since it is only 27.0% (0.270) pure, the mass = 24.7 g /0.270 = 91.5 g

Now using density, find the volume:

91.5 g x 1 ml/1.19 g = 76.9 ml