Calculate the percentage yield of sodium sulphate when 32.18 g of sulphuric acid reacts with excess sodium hydroxide to produce 37.91 g of sodium sulphate.
$H_{2} S O_{4} + 2 N a O H (a q) \to 2 H_{2} O (l) + N a_{2} S O_{4} (a q)$

91.37%

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80.9%

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78.13%

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61.35%

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Solution

The correct option is B 80.9%
The balanced equation is :
$H_{2} S O_{4} + 2 N a O H (a q) \to 2 H_{2} O (l) + N a_{2} S O_{4} (a q)$
Number of moles of $H_{2} S O_{4} = \frac{32.18 g}{98 g / m o l} = 0.33 m o l$
Number of moles of $N a_{2} S O_{4} = \frac{37.91 g}{142 g / m o l} = 0.27 m o l$
$H_{2} S O_{4}$ is the limiting reagent.
So, $N a_{2} S O_{4}$ formed will be $= \frac{0.33 m o l e o f H_{2} S O_{4} \times 1 m o l e o f N a_{2} S O_{4}}{1 m o l o f H_{2} S O_{4}} = 0.33 m o l$
Weight of $N a_{2} S O_{4} = 0.33 m o l \times 142 g / m o l = 46.86 g$
Percentage yield $= \frac{A c t u a l y i e l d}{t h e o r e t i c a l y i e l d} \times 100 = \frac{37.91}{46.86} \times 100 = 80.9 %$

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Similar questions

Q. Calculate the percentage yield of sodium sulphate when 32.18 g of sulphuric acid reacts with excess sodium hydroxide to produce 37.91 g of sodium sulphate.

H_{2} S O_{4} + 2 N a O H (a q) \to 2 H_{2} O (l) + N a_{2} S O_{4} (a q)

Q. Sulphuric acid reacts with sodium hydroxide as follows

H_{2} S O_{4}

2 N a O H

\to

N a_{2} S O_{4}

2 H_{2} O

When 1 L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is ?

Copper sulphate reacts with sodium hydroxide to form a blue precipitate of copper hydroxide and sodium sulphate. In an experiment, 15.95 g of copper sulphate reacted with 8.0 g of sodium hydroxide to form 9.75 g of copper hydroxide and 14.2 g of sodium sulphate. Which law of chemical combination is illustrated by this data?

Sulphuric acid reacts with sodium hydroxide as follows

$H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O$

When 1L of 0.1M sulphuric acid solutoin is allowed to react with 1L of 0.1M sodium hydroxide solutoin, the amount of sodium sulphate formed and its molarity in the solution obtained is ?

(a) $0.1 m o l L^{- 1}$ (b) $7.10 g$