What volume of oxygen is required to completely burn a mixture of 224 ml of methane and 112 ml of hydrogen at STP, to produce steam and carbon dioxide? If the steam formed is condensed to water, find the mass of water so formed.

602 ml, 1.44 g

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504 ml, 0.45 g

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800 ml, 2.54 g

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404 ml, 3.56 g

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Solution

The correct option is B 504 ml, 0.45 g
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

$2 H_{2} + O_{2} \to 2 H_{2} O$

Volumes of $O_{2}$ required for combustion of $C H_{4} = 224 \times 2 = 448 m l$

Volumes of $O_{2}$ required for combustion of $H_{2} = 112 \times \frac{1}{2} = 56 m l$

Total volume of $O_{2}$ required $= 448 + 56 = 504 m l$

Mass of water formed $= (\frac{448}{22400} + \frac{112}{22400}) \times 18 = 0.45 g m$

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What volume of oxygen is required to completely burn a mixture of 224 ml of methane and 112 ml of hydrogen at STP, to produce steam and carbon dioxide? If the steam formed is condensed to water, find the mass of water so formed.

The correct option is B 504 ml, 0.45 gCH4+2O2→CO2+2H2O2H2+O2→2H2OVolumes of O2 required for combustion of CH4=224×2=448 mlVolumes of O2 required for combustion of H2=112×12=56 mlTotal volume of O2 required=448+56=504 mlMass of water formed =(44822400+11222400)×18=0.45 gm