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%.
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