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Question

$$0.1$$ mole of $$H_{2}$$ and $$0.2\ mole$$ of $$I_{2}$$ are made to react in order to attain equilibrium $$H_{2} + I_{2}\rightleftharpoons 2HI$$. At equilirbrium $$20$$% $$H_{2}$$ is converted into $$HI$$, calculate amount of $$HI$$ formed.


Solution

$${ H }_{ 2 }+{ I }_{ 2 }\rightleftharpoons 2HI$$
$$0.1$$   $$0.2$$
at equilibrium 20% $${ H }_{ 2 }$$ is converted into $$HI$$. therefore, the amount of $$HI$$ is $$=2\times 0.1\times \dfrac { 20 }{ 100 } =0.04mol$$

Chemistry

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