What volume of gas will be released upon interaction of 250 g of concentrated nitric acid solution

| January 28, 2021 | education

What volume of gas will be released upon interaction of 250 g of concentrated nitric acid solution with a mass fraction of NHO3 in a solution of 33.3% with metallic sodium?

Let’s find the mass of НNO3 in solution.

W = m (substance): m (solution) × 100%,

hence m (substance) = (m (solution) × w): 100%.

m (substance) = (250 g × 33.3%): 100% = 83.25 g.

Let’s find the amount of substance НNO3.

n = m: M.

M (НNO3) = 63 g / mol.

n = 83.25 g: 63 g / mol = 1.32 mol.

Let’s find the quantitative ratios of substances.

8Na + 10HNO3conc = 8 NaNO3 + N2O + 5H2O.

For 10 mol of НNO3, there is 1 mol of N2O.

The substances are in quantitative ratios of 10: 1.

The amount of N2O substance is 10 times less than НNO3.

n (N2O) = 1/10 n (НNO3) = 1.32: 10 = 0.132 mol.

Let’s find the volume of N2O.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 0.132 mol × 22.4 L / mol = 2.96 L.

Answer: 2.96 liters.



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