Chalk weighing 400 g containing 10% of impurities reacts with 252 g of 15% nitric acid. Find the volume of released gas

| February 25, 2021 | education

Let’s find the mass of chalk without impurities.

100% – 10% = 90%

400 g – 100%,

X = 90%.

X = (400g × 90%): 100% = 360g.

Let’s find the amount of substance CaCO3.

M (CaCO3) = 100 g / mol.

n = 360 g: 100 g / mol = 3.6 mol.

Let’s find the mass of HNO3 in solution.

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

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

m (HNO3) = (252 g × 15%): 100% = 37.8 g.

Let’s find the amount of substance HNO3 by the formula:

n = m: M.

M (HNO3) = 63 g / mol.

n = 37.8 g: 63 g / mol = 0.6 mol (deficiency).

Let’s compose the reaction equation, find the quantitative ratios of substances.

CaCO3 + 2HNO3 = Ca (NO3) 2 + CO2 ↑ + H2O.

For 2 mol of HNO3 there is 1 mol of CO2. Substances are in quantitative ratios 2: 1. The amount of CO2 is 2 times less than the amount of HNO3.

n (CO2) = ½ n (HNO3) = 0.6: 2 = 0.3 mol.

Let’s find the volume of CO2.

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

V = 0.3 mol × 22.4 L / mol = 6.72 L.

Answer: 6.72 L



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