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Standard XII

Chemistry

Mole Concept

Calculate the...

Question

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. 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 ?

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) $7.10 g$

Q. If 2 moles of NaOH reacts with 4 moles of HCl to give sodium chloride and then the produced sodium chloride reacts with 3 moles of concentrated sulphuric acid to give sodium sulphate, find the total moles of sodium sulphate produced. The concerned reactions are

N a O H + H C l (d i l u t e d) \to N a C l + H_{2} O

2 N a C l + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H C l

Question 12

When sodium peroxide is treated with dilute sulphuric acid, we get .....

(a) sodium sulphate and water

(b) sodium sulphate and oxygen

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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?

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Mole Concept

Standard XII Chemistry

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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. H2SO4+2NaOH(aq)→2H2O(l)+Na2SO4(aq)

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