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10 mole of $SO_2$ and 15 mole of $O_2$ were passed over catalyst to produce 8 mole of $SO_3$.The ratio of $SO_2$ and $SO_3$ moles in mixture is

$\begin{array}{1 1}(a)\;\large\frac{5}{4}&(b)\;\large\frac{1}{4}\\(c)\;\large\frac{1}{2}&(d)\;\large\frac{3}{4}\end{array}$

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1 Answer

$SO_2+\large\frac{1}{2}$$O_2\iff SO_3$
Initial mole :$SO_2\Rightarrow 10,\large\frac{1}{2}$$O_2\Rightarrow 15$
Final mole :$SO_2\Rightarrow 10-x,\large\frac{1}{2}$$O_2\Rightarrow 15-\large\frac{x}{2},$$SO_3\Rightarrow x$
$x=8$
$\therefore SO_2: SO_3\Rightarrow 2 : 8$
$\Rightarrow 1 : 4$
Hence (b) is the correct answer.
answered Jan 8, 2014 by sreemathi.v
 

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