Question

if 50 ml of H2 and 30 ml of O2 react to form water vapour then the final volume of reaction mixture would be.?



pls...solve it fast...it's.urgent

Answer 1

Plz refer attachment below

Answer 2

Answer: 55 ml

Explanation: According to Avogadro's law, 1 mole of every gas contains $6.023\times 10^{23}$ particles and occupy 22.4 L at STP.

$2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$

From the balanced chemical chemical equation:

2 moles of  $H_2$ gas reacts with 1 mole of $O_2$ gas.

44800 mL of  $H_2$ gas reacts with 22400 mL of $O_2$ gas.

Thus 50 ml of $H_2$ gas will react with=$\frac{22400}{44800}\times 50=25mL$ of $O_2$ gas.

As hydrogen is the limiting reagent, and oxygen is the excess reagent.(30-25)=5 ml of oxygen will be left unused.

2 moles of $H_2$ produces 2 moles of water vapour $(H_2O)$

50 ml of  $H_2$ produces = 50 ml of water vapour $(H_2O)$

Also 5 ml of oxygen is left. Thus final volume of reaction mixture is (50+5)=55 ml.