A) 0.0075 & 0.147 moles
B) 0.0050 & 0.147 moles
C) 0.0075 & 0.347 moles
D) 0.0052 & 0.347 moles
Correct Answer: C
Solution :
[c] Moles of \[{{I}_{2}}\] taken \[=\frac{46}{254}=0.181\] |
\[\left( \because \text{no of moles}=\frac{\text{given mass}}{\text{molar mass}} \right)\] |
Moles of \[{{H}_{2}}\] taken \[=\frac{1}{2}=0.5\] |
Moles of \[{{I}_{2}}\] remaining \[=\frac{1.9}{254}=0.0075\] |
Moles of \[\]used \[=0.181-0.0075=0.1735\] |
Moles of \[{{\text{H}}_{\text{2}}}\] used \[=0.1735\] |
Moles of \[{{\text{H}}_{\text{2}}}\]remaining \[0.5-0.1735=0.3265\] |
Moles of HI formed \[=0.1735\times 2=0.347\] |
At equilibrium |
Moles of \[{{\text{I}}_{\text{2}}}=0.0075\] |
Moles of HI = 0.347 moles |
You need to login to perform this action.
You will be redirected in
3 sec