Determine how much hydrogen sulfide is released during the interaction of 300 g of sodium

| September 5, 2021 | education

Determine how much hydrogen sulfide is released during the interaction of 300 g of sodium chloride containing 22% impurities with concentrated sulfuric acid

Let us find the mass of NaCl without impurities.

100% – 22% = 78%.

300 g – 100%,

x g – 78%,

x = (300 g × 78%): 100% = 234 g.

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

n = m: M.

M (NaCl) = 58.5 g / mol.

n = 234 g: 58.5 g / mol = 4 mol.

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

2 NaCl + H2SO4 = Na2SO4 + 2 HCl ↑.

According to the reaction equation, 2 mol of NaCl accounts for 2 mol of HCl. Substances are in quantitative ratios 1: 1.

n (HCl) = n (NaCl) = 4 mol.

Find the volume of HCl.

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

V = 4 mol × 22.4 L / mol = 89.6 L.

Answer: 89.6 l.



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