Calculate the volume of H2 that is released during the interaction of 13.8 g glycerin with metallic potassium – 11.7 g.

| September 4, 2021 | education

Let’s find the amount of the substance potassium and glycerin.

n = m: M.

M (K) = 39 g / mol.

n = 11.7 g: 39 g / mol = 0.3 mol.

M (C3H5 (OH) 3) = 92g / mol.

n = 13.8: 92 g / mol = 0.15 mol.

Let’s find the quantitative ratios of substances.

2C3H5 (OH) 3 + 6K = 2C3H5 (OK) 3 + 3H2 ↑.

Glycerin is given in excess, since glycerol and potassium are in a ratio of 1: 3, the amount of potassium substance should be 0.15 × 3 = 0.45 mol.

We solve the problem by potassium.

For 6 mol K, there is 3 mol of H2.

Substances are in quantitative ratios 6: 3 = 2: 1.

The amount of hydrogen substance will be 2 times less than K.

n (H2) = ½ n (K) = 0.3: 2 = 0.15 mol.

Let’s find the volume of H2.

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 = 0.15 mol × 22.4 L / mol = 3.36 L.

Answer: 3.36 liters.



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