A) 5 g
B) 9 g
C) 10 g
D) 12 g
Correct Answer: B
Solution :
n \[=\frac{16.8}{22.4}=0.75\]mole of \[{{H}_{2}}\]and \[{{O}_{2}}\] \[2{{H}_{2}}O\to \underset{2:1}{\mathop{2{{H}_{2}}+{{O}_{2}}}}\,\]You need to login to perform this action.
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