A) 2 mol
B) 3 mol
C) 4 mol
D) 1 mol
Correct Answer: C
Solution :
Key Idea (i)\[\to \] (ii) Amount of water produced is decided by limited reactant (ie, the reactant which is used in small amount) \[\to \] \[\to \] \[\to \] \[\to \]\[\to \] \[\text{N}{{\text{a}}^{\text{+}}}\]\[C{{O}_{2}}\]gives = 1 mol\[=1\times 12=12\text{kg}\,\text{m}{{\text{s}}^{-1}}\] \[=2\times 8=16\text{kg}\,\text{m}{{\text{s}}^{-1}}\]\[\therefore \]will give\[=\sqrt{{{(12)}^{2}}+{{(16)}^{2}}}=20\text{kg}\,\text{m}{{\text{s}}^{-1}}\]mol.
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