$24 m l$ of water gas containing only hydrogen and carbon monoxide in equal proportions by volume, are exploded with $80 m l$ of air, $20$ by volume of which is oxygen. If all gaseous volumes are measured at room temperature and pressure, calculate the composition by volume of the resulting gaseous mixture.

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Solution

$2 H_{2} + O_{2} \to 2 H_{2} O$
$2 V$
$2 C O + O_{2} \to 2 C O_{2}$
$2 V$
Given that
$2 V_{p / 2} = 2 V_{c o}$
So, $4 V = 24$
$V = \frac{24}{8} = 4 m^{2}$
Volume of air $= \frac{20 \times 80}{100} = 16$
$2 V H_{2}$ react with $V L$ of $O_{2}$
So, $4 m L O_{2}$ need $2 \times 4 = 8 m L H_{2}$
$16 m L O_{2}$ needed $= \frac{16 \times 8}{4} = 32 H_{2}$
So, $H_{2} = C O = 32 m L$
$O_{2} = 16 m L$

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24ml of water gas containing only hydrogen and carbon monoxide in equal proportions by volume, are exploded with 80ml of air, 20 by volume of which is oxygen. If all gaseous volumes are measured at room temperature and pressure, calculate the composition by volume of the resulting gaseous mixture.

2H2+O2→2H2O2V2CO+O2→2CO22VGiven that2Vp/2=2VcoSo, 4V=24V=248=4m2Volume of air =20×80100=162VH2 react with VL of O2So, 4mLO2 need 2×4=8mLH216mLO2 needed =16×84=32H2So, H2=CO=32mLO2=16mL