a. 43.1 g
b. 38.2%
<h3>Further explanation</h3>
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%
Is that chemistry?
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Explanation:
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Answer:
Explanation:
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