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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

$2, 11$

Vedclass · Answer

(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?

Similar Questions

What ratio by mass of $Ne$ and $CH_4$ should be mixed so that the partial pressure exerted by each gas is the same?

What quantity of ammonium sulphate is necessary for the production of $NH_3$ gas sufficient to neutralize a solution containing $292 \ g$ of $HCl$? $(HCl=36.5, (NH_4)_2SO_4=132, NH_3=17)$

$4.0 \, g$ of caustic soda is dissolved in $100 \, mL$ of solution. The normality of the solution is: (in $, N$)

$0.53 \ g$ of $Na_2CO_3$ has been dissolved in $100 \ mL$ of a sodium carbonate solution. The normality of the solution will be

The normality of orthophosphoric acid having purity of $70\%$ by weight and specific gravity $1.54$ would be ............. $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module