A) \[3\,\,mol\]
B) \[4\,\,mol\]
C) \[1\,\,mol\]
D) \[2\,\,mol\]
Correct Answer: B
Solution :
Moles of\[{{H}_{2}}=\frac{10}{2}=5\,\,mol\] Moles of\[{{O}_{2}}=\frac{64}{32}=2\,\,mol\] \[\underset{(5\,\,mol)}{\mathop{2{{H}_{2}}}}\,+\underset{(2\,\,mol)}{\mathop{{{O}_{2}}}}\,\xrightarrow{{}}\underset{(4\,\,mol)}{\mathop{2{{H}_{2}}O}}\,\] Here \[{{O}_{2}}\] is limiting reagent. So, amount of product, \[{{H}_{2}}O\] obtained depends on the amount of\[{{O}_{2}}\]. According to equation, 1 mole \[{{O}_{2}}\] on reaction with \[{{H}_{2}}\] gives\[=2\] mole of\[{{H}_{2}}O\] 2 mole \[{{O}_{2}}\] on reaction with\[{{H}_{2}}\] will give \[=\frac{2\times 2}{1}=4\,\,mole\]of\[{{H}_{2}}O\] Note 1 mole of \[{{H}_{2}}\] is in excess.
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