# What mass of nitric acid reacted with zinc oxide if you got 400 g of zinc nitrate. And the reaction yield is 90%

October 15, 2021 | education

| Let’s write the equation:

2HNO3 + ZnO = Zn (NO3) 2 + H2O – ion exchange reaction, zinc nitrate is released;

2 mol 1 mol;

Let’s make calculations using the formulas:

M (HNO3) = 63 g / mol;

M Zn (NO3) 2 = 189.3 g / mol;

Determine the amount of zinc nitrate moles:

Y Zn (NO3) 2 = m / M = 400 / 189.3 = 2.1 mol;

Let’s make the proportion:

X mol (HNO3) – 2.1 mol Zn (NO3) 2;

-2 mol -1 mol hence, X mol (HNO3) = 2 * 2.1 / 1 = 4.2 mol;

We find the mass of nitric acid (theoretical):

m (HNO3) = Y * M = 4.2 * 63 = 264.6 g.

Let’s calculate the practical mass, taking into account the product yield of 90%:

m (HNO3) = 0.9 * 264.6 = 238.14 g.

Answer: the mass of nitric acid is 238.6 g.