Question

The combustion of hydrogen-oxygen mixture is used to produce water at very high temperature $\left(\approx {2500}^{\circ }\text{C}\right)$ needed for certain types of welding operations.

${\text{H}}_{\text{2}}\left(\text{g}\right)+\frac{1}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{g}\right);$ $\Delta \text{H}=-241.8\text{kJ}$

Quantity of heat (in kJ) evolved when a 180 g mixture containing equal parts of ${\text{H}}_{\text{2}}\text{and}{\text{O}}_{\text{2}}$ mass burned, is:

• A. $\text{- 2}\text{.72}\times \text{1}{\text{0}}^3\text{kJ}$
• B. $\text{- 6}\text{.80}\times \text{1}{\text{0}}^2\text{kJ}$
• C. $\text{- 1}\text{.36}\times \text{1}{\text{0}}^3\text{kJ}$
• D. $\text{- 4}\text{.3524}\times \text{1}{\text{0}}^4\text{kJ}$

Answer 1

Answer: (C)

$\text{- 1}\text{.36}\times \text{1}{\text{0}}^3\text{kJ}$

Mass of mixture = 180 g

Mass of $H_2$ = 90 g

Mass of $O_2$ = 90 g

2g of hydrogen react with 18g of oxygen as per the balanced reaction

So, $O_2$ is limiting reagent,

Since 16 g of $O_2$ produce heat = -241.8 kJ

Heat produce by 90 g of $O_2$ = $\frac{-241.8}{16}\times 90=-1360kJ$