When 10 mol of SO_2 and 15 mol of O_2 are m | vedclass

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Question

(A) The balanced chemical equation is: $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$.Initial moles: $SO_2 = 10 \ mol$,$O_2 = 15 \ mol$,$SO_3 = 0 \ mol$.Acc

Vedclass · Accepted Answer

$2, 11$

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(A) The balanced chemical equation is: $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$.Initial moles: $SO_2 = 10 \ mol$,$O_2 = 15 \ mol$,$SO_3 = 0 \ mol$.According to the stoichiometry,$2 \ mol$ of $SO_2$ produces $2 \ mol$ of $SO_3$.Since $8 \ mol$ of $SO_3$ is formed,the amount of $SO_2$ consumed is $8 \ mol$.Remaining $SO_2 = 10 - 8 = 2 \ mol$.For $2 \ mol$ of $SO_2$,$1 \ mol$ of $O_2$ is required.So,for $8 \ mol$ of $SO_2$,the amount of $O_2$ consumed is $8 / 2 = 4 \ mol$.Remaining $O_2 = 15 - 4 = 11 \ mol$.Thus,the unreacted moles are $2 \ mol$ of $SO_2$ and $11 \ mol$ of $O_2$.

When $10 \ mol$ of $SO_2$ and $15 \ mol$ of $O_2$ are mixed and passed over a catalyst,$8 \ mol$ of $SO_3$ is formed. How many moles of $SO_2$ and $O_2$ remain unreacted?

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