Determine the mass and amount of the formed salt upon the interaction of 0.2 mol of sodium hydroxide

| July 4, 2021 | education

Determine the mass and amount of the formed salt upon the interaction of 0.2 mol of sodium hydroxide with 12 liters of carbon dioxide.

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

n = V: Vn.

n (CO2) = 12 L: 22.4 L / mol = 0.54 mol.

Carbon dioxide is given in excess.

The amount of the salt substance is calculated by the deficiency (sodium hydroxide).

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

2NaOH + CO2 = Na2CO3 + H2O.

According to the reaction equation, there is 1 mole of sodium carbonate for 2 mol of sodium hydroxide. Substances are in quantitative ratios of 2: 1.

The amount of the salt substance will be 2 times less than the amount of the sodium hydroxide substance.

n (Na2CO3) = 1 / 2n (NaOH) = 0.2: 2 = 0.1 mol.

Find the mass of sodium carbonate by the formula:

m = n × M,

M (Na2CO3) = 23 × 2 + 12 + 48 = 106 g / mol.

m = 0.1 mol × 106 g / mol = 10.6 g.

Answer: 10.6 g.



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