What is the mass of sodium carbonate obtained by passing carbon monoxide (IV) with a mass of 22 g through a solution

What is the mass of sodium carbonate obtained by passing carbon monoxide (IV) with a mass of 22 g through a solution with a mass of 20 g with a mass fraction of sodium hydroxide of 25%?

Given:

m (CO2) = 22 g

m (NaOH) = 20 g

w% (NaOH) = 25%

To find:

m (Na2CO3) -?

Decision:

1) We compose the reaction equation according to the condition of the problem:

2NaOH + CO2 = Na2CO3 + H2;

2) Find the mass of alkali in the solution:

m (NaOH) = 20 g * 0.25 = 5 g;

3) Find the amount of carbon monoxide and sodium hydroxide:

n (CO2) = m: M = 22 g: 44 g / mol = 0.5 mol

n (NaOH) = m: M = 5 g: 40 g / mol = 0.125 mol

We start from a lower value to get more accurate calculations. We work with NaOH:

4) We compose logical equality:

if 2 mol NaOH gives 1 mol Na2CO3,

then 0.125 mol NaOH will give x mol Na2CO3,

then x = 0.0625 mol.

5) Find the mass of sodium carbonate formed as a result of the reaction:

m (Na2CO3) = n * M = 0.0625 mol * 106 g / mol = 6.625 g;

Answer: m (Na2CO3) = 6.625 g.