A) 276 g
B) 345 g
C) 690 g
D) 1380 g
Correct Answer: B
Solution :
[b] \[A{{g}_{2}}C{{O}_{3}}(s)\to 2Ag(s)+C{{O}_{2}}(g)+1/2{{O}_{2}}(g)\] |
\[{{C}_{2}}{{H}_{2}}+5/2{{O}_{2}}\to 2C{{O}_{2}}+{{H}_{2}}O\] |
By Stoichiometry of reaction |
Moles of \[C{{O}_{2}}\] formed\[=\frac{11.2}{22.4}=\frac{1}{2}\] |
Moles of \[{{O}_{2}}\] required \[=\frac{5}{4}\times \frac{1}{2}=\frac{5}{8}\] |
\[\therefore \] Moles of \[A{{g}_{2}}C{{O}_{3}}\] required \[=2\times \frac{5}{8}=\frac{5}{4}\] |
Mass of \[A{{g}_{2}}C{{O}_{3}}\] required \[=\frac{5}{4}\times 276=345g\] |
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