Determine the mass of salt that is formed by the interaction of 3.65 g of HCL with 10.6 g of Na2Co3

| May 21, 2021 | education

Determine the mass of salt that is formed by the interaction of 3.65 g of HCL with 10.6 g of Na2Co3. What V of gas will be released in this case.

m (HCl) = 3.65 g.

m (Na2CO3) = 10.6 g.

Let us determine the mass of the formed salt and the volume of the gas. We write down the reaction equation.

2HCL + Na2CO3 = 2NaCl + H2O + CO2

First, the amount of substance perchloric acid and sodium carbonate.

n = m / M.

M (HCl) = 35.5 + 1 = 36.5 g / mol.

M (Na2CO3) = 23 * 2 + 12 + 16 * 3 = 106 g / mol.

n (HCl) = 3.65 / 36.5 = 0.1 mol.

n (Na2CO3) = 10.6 / 106 = 0.1 mol.

This means that there is a lack of perchloric acid.

We find by the equation of reactions the mass of salt.

0.1 mol HCl – x mol NaCl

2 mol HCl – 1 mol NaCl

X = 0.1 * 1: 2 = 0.05 mol NaCl.

m (NaCl) = 0.05 * 58.5 = 2.925 g.

We find the volume of gas.

0.1 mol НСl – x mol СО2

2 mol HCl – 1 mol CO2

X = 0.1 * 1: 2 = 0.05 mol.

n = V / Vm.

V (CO2) = 0.05 * 22.4 = 1.12 liters.

Answer: m (NaCl) = 2.925 g. V (CO2) = 1.12 l.



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