Question

1. 32.5grams of NaOH (sodium hydroxide) reacts an excess of CO2 (carbon dioxide).

Answer 1

a. 43.1 g

b. 38.2%

Further explanation

Given

32.5 grams of NaOH

Required

The theoretical yield of Na₂CO₃

The percent yield

Solution

Reaction

2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l)

mol NaOH :

= mass : MW

= 32.5 : 40 g/mol

= 0.8125

mol Na₂CO₃ from the equation :

= 1/2 x mol NaOH

= 1/2 x 0.8125

= 0.40625

a.

Mass Na₂CO₃ :

= mol x MW Na₂CO₃

= 0.40625 x 106 g/mol

= 43.0625≈43.1 g

b. % yield = (actual/theoretical) x 100%

%yield = 16.45/43.1 x 1005

%yield = 38.17%≈38.2%