calculate the amount of water(H2O) formed during the complete combustion of 96g methane(CH4)(stoichiometric calculation)???
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CH4+2O2 ⇒CO2+2H2O
1 mol CH4 =2 mol H2O
96 g CH4
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no. of moles = Given mass ÷ Gram molecular mass[GM/GMM]
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GM = 96 g
(C=12,H=1)
GMM=12×1+1×4=16u
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no. of moles⇒96÷16
⇒6 mol
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6 mol CH4 =2×6 mol H2O
=12 mol H2O
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no. of moles⇒GM/GMM
12⇒GM/18
GM =18 ×12
= 216 g
amount of water(H2O)⇒216 g
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