A football bladder contains an equimolar proportion of and gases. What will be the composition by ma — States of Matter and Gaseous State Chemistry Question
Question
A football bladder contains an equimolar proportion of $H_2$ and $O_2$ gases. What will be the composition by mass of the mixture effusing out initially if the football is punctured? (Report as $H_2 : O_2$).
💡 Solution & Explanation
The initial ratio of moles effusing is $r_{H_2}/r_{O_2} = n_{H_2\_effused} / n_{O_2\_effused} = (P_{H_2}/P_{O_2}) \times \sqrt{M_{O_2}/M_{H_2}}$. Since the mixture is equimolar, partial pressures are equal. Thus, $n_{H_2}/n_{O_2} = \sqrt{32/2} = 4:1$. To find the mass ratio, multiply by molar mass: $(4 \text{ moles} \times 2\text{ g/mol}) / (1 \text{ mole} \times 32\text{ g/mol}) = 8 / 32 = 1 : 4$.