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

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.

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