# The reaction of 80 g of copper oxide with nitric acid yielded 141 g of copper sodium (2).

**The reaction of 80 g of copper oxide with nitric acid yielded 141 g of copper sodium (2). Calculate the mass fraction of the salt yield**

Let’s find the amount of substance CuO.

n = m: M.

M (CuO) = 80 g / mol.

n = 80 g: 80 g / mol = 1 mol.

Let’s find the quantitative ratios of substances.

CuO + 2HNO3 = Cu (NO3) 2 + H2O.

For 1 mol of CuO, there is 1 mol of Cu (NO3) 2.

Substances are in quantitative ratios of 1: 1.

The amount of substance will be the same.

n (Cu (NO3) 2) = n (CuO) = 1 mol.

Let us find the mass of Cu (NO3) 2.

M (Cu (NO3) 2) = 188 g / mol.

m = n × M.

m = 188 g / mol × 1 mol = 188 g.

188 g received according to theory (according to calculations). According to the condition of the problem, the yield of Cu (NO3) 2 was 141 g.

188 g – 100%,

141 g – x%.

x = (141g × 100%): 188g = 75%.

Answer: 75%.