How many grams of sodium nitrate is needed to react with the end of sulfuric acid

| January 31, 2021 | education

How many grams of sodium nitrate is needed to react with the end of sulfuric acid, which, as a result of the reaction, get 180 g of nitric acid with a yield of 90 percent

Let’s implement the solution:

According to the condition of the problem, we write the data:
X g -?
m = 180 g; W = 90%

2NaNO3 + H2SO4 (conc.) = Na2SO4 + 2HNO3 – ion exchange, nitric acid obtained;

Calculations:
M (NaNO3) = 84.9 g / mol;

M (HNO3) = 63 g / mol.

We find the mass of the acid, the original substance:
W = m (practical) / m (theoretical) * 100;

m (HNO3) = 180 / 0.90 = 200 g (theoretical weight);

Y (HNO3) = m / M = 200/63 = 3.2 mol;

Y (NaNO3) = 3.2 mol since the amount of substances according to the equation is 2 mol.

Find the mass of salt:
m (NaNO3) = Y * M = 3.2 * 84.9 = 271.68 g

Answer: the mass of sodium nitrate is 271.68 g



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