A) 2 mol
B) 3 mol
C) 4 mol
D) 1 mol
Correct Answer: C
Solution :
| Key Idea (i)\[{{H}_{2}}+\frac{1}{2}{{O}_{2}}\xrightarrow{{}}{{H}_{2}}O\] |
| (ii) Amount of water produced is decided by limited reactant (ie, the reactant which is used in small amount) \[{{H}_{2}}+\frac{1}{2}{{O}_{2}}\xrightarrow{{}}{{H}_{2}}O\] |
| \[1\,\,mol\,\frac{1}{2}\,mol\,\,1\,mol\] |
| \[\frac{10}{2}\,mol\,\frac{64}{32}\,mol\,?\] |
| \[=5\,mol\,\,=2\,mol\] |
| \[\text{N}{{\text{a}}^{\text{+}}}\] \[\frac{1}{2}\,mol\,\,{{O}_{2}}\] gives \[=1\,\,mol\,\,{{H}_{2}}O\] |
| \[=2\times 8=16\text{kg}\,\text{m}{{\text{s}}^{-1}}\] \[2\,mol\,{{O}_{2}}\] will give\[=1\,\times \,2\times \,2\,=\,4\,mol\]. |
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