Question

The combustion of hydrogen-oxygen mixture is used to produce very high temperatures $(\approx {2500}^{o}C)$ needed for certain types of welding operations.
$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(g\right);△H=-230.9kJ$
Quantity of heat, (in kJ) evolved when a 200 g mixture containing equal parts of $H_2$ and $O_2$ by mass is burned, is

• A. $2.72\times {10}^{3}kJ$
• B. $1.36\times {10}^{3}kJ$
• C. $1.44\times {10}^{3}kJ$
• D. $4.3524\times {10}^{4}kJ$

Answer 1

The correct option is C $1.44\times {10}^{3}kJ$

Total mass taken = 200 g
So, Mass of each $O_{2}and{H}_2$taken
= $\frac{200}{2}=100g$
Moles of $O_2$ taken $=100g=\frac{100}{32}=3.125$
Moles of $H_2$ taken $=100g=\frac{100}{2}=50.00mol$
Clearly from here we can say that,
$O_2$ is the limiting reagent.
$H_2$ used in combistion = $2\times molesof{O}_2=6.25mol$
Thus, heat evolved $=Heatevolvedforonemole\times$ No. of moles of $H_2$ involved
$=-230.9kJmo{l}^{-1}\times 6.25mol$
$=-1443.125kJ=-1.44\times {10}^{3}kJ$