Question
$$10$$ $$moles$$ $$S{O_2}$$ and $$15$$ $$moles$$ $${O_2}$$ were allowed to react over a suitable catalyst. $$8$$ $$moles$$ of $$S{O_3}$$ were formed. The remaining moles of $$S{O_2}$$ and $${O_2}$$ respectively are -
A.
$$2\,moles,\,11\,moles$$
B.
$$2\,moles,\,8\,moles$$
C.
$$4\,moles,\,5\,moles$$
D.
$$8\,moles,\,2\,moles$$
Answer :
$$2\,moles,\,11\,moles$$
Solution :
$$\eqalign{
& 2S{O_2} + \,\,\,\,\,\,{O_2} \to \,\,\,2S{O_3} \cr
& 10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr
& 10 - 2x\,\,\,\,\,15 - x\,\,\,\,\,\,2x \cr
& \because \,\,2x = 8\,\,\,x = 4 \cr
& {\text{Hence, remaining,}}\,\,S{O_2} = 10 - 8 = 2\,moles, \cr
& {O_2} = 15 - 4 = 11\,moles \cr} $$