A) 3mol
B) 4mol
C) 1 mol
D) 2mol
Correct Answer: D
Solution :
Moles of\[{{H}_{2}}=\frac{10}{2}=5\,\text{mol}\] Moles of \[{{O}_{2}}=\frac{64}{32}=2\,\text{mol}\] \[\underset{(5\,mol)}{\mathop{2{{H}_{2}}}}\,+\underset{(2\,mol)}{\mathop{{{O}_{2}}}}\,\xrightarrow{{}}\underset{(4\,mol)}{\mathop{2{{H}_{2}}O}}\,\] Here \[{{\text{O}}_{\text{2}}}\] is limiting reagent. So, amount of product, \[{{\text{H}}_{\text{2}}}\text{O}\]obtained depends on the amount of \[{{\text{O}}_{\text{2}}}\text{.}\] 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 \[=\frac{2\times 2}{1}=4\text{mole}\,\text{of}\] Note 1 mole of H2 is in excess.
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