Question

calculate the amount of water(H2O) formed during the complete combustion of 96g methane(CH4)(stoichiometric calculation)???


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Answer 1

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.

Answer 2

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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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