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  4. 2 moles of H 2 S and 11.2 L of SO 2 at N.T.P. reacts to form x moles of sulphur. The value of x is:

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Q. 2 moles of $H _{2} S$ and $11.2\, L$ of $SO _{2}$ at $N.T.P.$ reacts to form x moles of sulphur. The value of $x$ is:

Some Basic Concepts of Chemistry

Solution:

$22.4 \,L$ of $SO _{2}$ at $N \cdot T.P =1$ moles

$11.2\, L$ of $SO _{2}$ at $N . T.P =1 / 2$ moles

$SO _{2}+2 H _{2} S \rightarrow 3 S +2 H _{2} O$

Here $SO _{2}$ is a limiting reactant

$\therefore $ moles of sulphur produced $=3 \times$ moles of $SO _{2}$

$=1.5$ moles