A) 100
B) 152
C) 50
D) 5
Correct Answer: B
Solution :
\[\underset{\begin{smallmatrix} 1\,mol \\ 0.95\,mol \end{smallmatrix}}{\mathop{{{C}_{8}}{{H}_{18}}}}\,+8.5{{O}_{2}}\xrightarrow[{}]{{}}\underset{\begin{smallmatrix} 8\,mol \\ 7.6mol \end{smallmatrix}}{\mathop{8CO}}\,+9{{H}_{2}}O\] Emitted gas contains 0.05 mol \[{{C}_{8}}{{H}_{18}}\] and 7.6 mol of CO, if 1 mol of \[{{C}_{8}}{{H}_{18}}\] is the fuel. \[\frac{CO}{{{C}_{8}}{{H}_{18}}}=\frac{7.6}{0.05}=152.\]You need to login to perform this action.
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