A) 3 mol
B) 4 mol
C) 1 mol
D) 2 mol
Correct Answer: B
Solution :
\[\underset{1\,mol}{\mathop{{{H}_{2}}}}\,+\underset{0.5\,mol}{\mathop{\frac{1}{2}{{O}_{2}}}}\,\xrightarrow[{}]{{}}\underset{1\,mol}{\mathop{{{H}_{2}}O}}\,\] \[\frac{10}{2}=5mol\frac{64}{32}=2\,mol\,4mol\] Because oxygen is the limiting reagent, hence 4 moles of\[{{H}_{2}}O\]formed.You need to login to perform this action.
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