Gas was released during the interaction of sodium carbonate weighing 132.5 g with hydrochloric acid.

| June 22, 2021 | education

Gas was released during the interaction of sodium carbonate weighing 132.5 g with hydrochloric acid. Find the mass of the resulting salt.

To solve this problem, we write down given: m (Na2CO3) = 132.5 g, this salt interacts with hydrochloric acid (HCl).
Find: mass of salt.
Decision:
Let’s write down the reaction equation.
Na2CO3 + 2HCl = 2NaCl + H2O + CO2
The salt, the mass of which we need to calculate, has the formula (NaCl)
Calculate the molar mass of Na2CO3 and NaCl
M (Na2CO3) = 23 * 2 + 12 + 16 * 3 = 106 g / mol
M (NaCl) = 23 + 35.5 = 58.5 g / mol
Since sodium chloride formed 2 mol, we need to multiply the molar mass by the number of moles and get 117 g.
Over Na2CO3 we sign 132.5 g, and under Na2CO3 – 106 g. Over NaCl we sign x g, and under NaCl – 117 g.
Let’s compose and solve the proportion
x = 117 * 132.5 / 106 = 146.25 g
Answer: 146.25 g



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