Looking for classes? Ksquare Career Institute, Bengaluru →
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be
A
1 mol
B
2 mol
C
3 mol
D
4 mol
Detailed Solution
Reaction: $2H_2 + O_2 \rightarrow 2H_2O$
Moles of $H_2$ = $\frac{10}{2} = 5$ mol; moles of $O_2$ = $\frac{64}{32} = 2$ mol
2 mol of $O_2$ needs $2\times2 = 4$ mol of $H_2$; 5 mol of $H_2$ is available, so $H_2$ is in excess and $O_2$ is the limiting reagent.
1 mol of $O_2$ gives 2 mol of $H_2O$, so 2 mol of $O_2$ gives $2\times2 = 4$ mol of $H_2O$.
Amount of water produced = 4 mol (1 mol of $H_2$ is left over)
Moles of $H_2$ = $\frac{10}{2} = 5$ mol; moles of $O_2$ = $\frac{64}{32} = 2$ mol
2 mol of $O_2$ needs $2\times2 = 4$ mol of $H_2$; 5 mol of $H_2$ is available, so $H_2$ is in excess and $O_2$ is the limiting reagent.
1 mol of $O_2$ gives 2 mol of $H_2O$, so 2 mol of $O_2$ gives $2\times2 = 4$ mol of $H_2O$.
Amount of water produced = 4 mol (1 mol of $H_2$ is left over)