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Standard XII

Chemistry

Stoichiometric Calculations

SO2Cl2 sulphu...

Question

$S O_{2} C l_{2}$ (sulphuryl chloride) reacts with water to give a mixture of $H_{2} S O_{4}$ and $H C l$ . What volume of $0.2 M B a (O H)_{2}$ is needed to completely neutralize $25$ mL of $0.2 M S O_{2} C l_{2}$ solution.

25

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50

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100

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200

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Solution

The correct option is C $50$ mL
The reaction is as follows:
$S O_{2} C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l$
Millimoles of $H_{2} S O_{4}$ produced $= 5$
Millimoles of $H C l$ produced $= 10$
Millimoles of $B a (O H)_{2}$ required $= 5 (f o r H_{2} S O_{4}) + \frac{10}{2} (f o r H C l) = 10$
$∴ M \times V = 10$
or, $0.2 \times V = 10$
Therefore, $V = 50$ mL

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SO2Cl2 (sulphuryl chloride) reacts with water to give a mixture of H2SO4 and HCl. What volume of 0.2MBa(OH)2 is needed to completely neutralize 25 mL of 0.2MSO2Cl2 solution.

The correct option is C 50 mLThe reaction is as follows:SO2Cl2+2H2O→H2SO4+2HClMillimoles of H2SO4 produced =5Millimoles of HCl produced =10Millimoles of Ba(OH)2 required =5(forH2SO4)+102(forHCl)=10∴M×V=10 or, 0.2×V=10Therefore, V=50 mL

$S O_{2} C l_{2}$ (sulphuryl chloride) reacts with water to give a mixture of $H_{2} S O_{4}$ and $H C l$ . What volume of $0.2 M B a (O H)_{2}$ is needed to completely neutralize $25$ mL of $0.2 M S O_{2} C l_{2}$ solution.