calculate the amount of water(H2O) formed during the complete combustion of 96g methane(CH4)(stoichiometric calculation)???
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Answer:
CH4 (g)+2O2 (g)————-> CO2 (g) +2H2O (g)
Now according to the reaction 1 mole of CH4 gives 2 moles of H2O
And 16 grams of Methane mean 1 mole (moles=given mass/molecular mass. Molecular mass of methane is 16)
so 1 mole of methane = 16/16 (given mass 16 and molecular mass 16)
since 1 mole of methane is getting used up to form 2 mole of water.
number of moles=given mass/molecular mass ——- eq. 1
molecular mass of water is 18.
2=x/18 ( where x is the mass of water produced , using equation 1)
2*18=36grams
so 36 grams of water will be produced.
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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