You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stoichiometric Calculations

The weight of...

Question

The weight of a mixture containing $H C l$ and $H_{2} S O_{4}$ is $0.1 g$ . On treatment with an excess of an $A g N O_{3}$ solution, the acid mixture gives $0.1435 g$ of $A g C l$ . Weight percentage of the $H_{2} S O_{4}$ in the mixture is: (Molecular mass of $H, C l, S, O, A g, N$ are $1 u, 35.5 u, 32 u, 16 u 108 u, and 14 u$ )

Open in App

Solution

$A g N O_{3}$ reacts with $H C l$ to give $A g C l$ .
$A g N O_{3} + H C l \to A g C l + H N O_{3}$
Given, mass of $A g C l = 0.1435 g$
$\Rightarrow m o l e s o f A g C l = \frac{0.1435}{143.5} = 10^{- 3} m o l$
Accroding to the stoichiometry,
Moles of $A g C l$ = moles of $H C l$
Hence Moles of $H C l = 10^{- 3} m o l$
Mass of $H C l = 36.5 \times 10^{- 3} = 0.0365 g$
So, mass of $H_{2} S O_{4} = (0.1 - 0.0365) g = 0.0635 g$
$weight % of H_{2} S O_{4} = \frac{.0635 \times 100}{0.1} = 63.5 %$

Suggest Corrections

Similar questions

H_{2} S O_{4}

0.1

A g N O_{3}

0.1435

H_{2} S O_{4}

H_{2} S O_{4}

A g N O_{3}

H_{2} S O_{4}

H_{2} S O_{4}

A g N O_{3}

H_{2} S O_{4}

H C l

H_{2} S O_{4}

10

1 N

N a O H

20

A g N O_{3}

0.1435

A g C l

H C l

H_{2} S O_{4}

Join BYJU'S Learning Program