2H\(_2\) + O\(_2\) → 2H\(_2\)O

From the equation above, calculate the volume of unreacted oxygen gas if a mixture of 50cm\(^3\) of hydroden and 75cm\(^3\) of oxygen are involved

Derek644222

2 years ago

first we need to find the limiting reactant because when one of the reactants in a chemical reaction is in EXCESS, it enables the other reactant to be used up completely.
so from the equation above we have 50cm^3 of Hydrogen and 75cm^3 of oxygen
Now to determine the reactant in excess:
suppose Hydrogen (H2) is completely used-up Hence from the equation of reaction
2H2 + O2 → 2H2O
2 vol of H2 require = 1 vol of O2
:.50 vol of H2 require = (1 x 50)/2
= 25 vol(cm^3) of O2
since the amount of O2 available (75cm^3) is more than the amount required (25cm^3) then, O2 is in excess by (75-25) cm^3 = 50cm^3
thus making 50cm^3 volume of unreacted oxygen
hence making option B correct

RiverHeights

2 years ago

The wording for this question is wild