With the complete dissolution of 3.2 g of copper in concentrated nitric acid, 3 g of nitric oxide were released (4).

With the complete dissolution of 3.2 g of copper in concentrated nitric acid, 3 g of nitric oxide were released (4). What is the mass fraction of the yield of nitric oxide (4) from the theoretically possible?

Let’s write the reaction equation:
Cu + 4HNO3 = Cu (NO3) 2 + 2NO2 ↑ + 2H2O.
It can be seen from the reaction equation that:
ν (NO2) / 2 = ν (Cu).
m (Cu) / M (Cu) = m (NO2) / (2 * M (NO2)).
Let’s define the molar masses:
M (NO2) = 14 + 16 * 2 = 46 g / mol.
M (Cu) = 65.5 g / mol.
Determine the mass of nitric oxide (IV), which theoretically should be released during the reaction:
m (NO2) = m (Cu) * 2 * M (NO2) / M (Cu).
Substitute the numerical values:
m (NO2) = m (Cu) * 2 * M (NO2) / M (Cu) = 3.2 * 2 * 46 / 65.5 = 4.63 g
Let’s make the proportion:
4.63 – 100%
3 – x, hence:
x = 3 * 100 / 4.63 = 65%
Answer: the mass fraction of the yield of nitrogen oxide (IV) is 65%.