Question

Calculate the amount of H2O in gram produced by combustion of 36 gram of Methane

Answer 1

Answer:

2.70x10^24

Explanation:

CH4 + 2O2 ------------ CO2 + 2H2O

Moles = mass/molar mass, Moles = N/Avogadro's constant

Moles of CH4=36g/16gmol-1

Moles of H2O = N/6.022x10^23

N/6.022x10^23 X 16/36 = 2/1

N = 2 x 36 x 6.022x10^23/16

N=2.70x10^24

Answer 2

Answer:

81 grams of water.

Explanation:

Weight of CH4(methane)=12+(4×1)=16 g

Weight of H2O(water)=1+(2×16)=18 g

COMBUSTION OF METHANE(CH4):

CH4 + 2 O2 → CO2 + 2 H2O

So, clearly one mole of methane gives 2 moles of water.

Given that,36 g of methane is involved in reaction.

No. of moles of methane=Given mass/Molecular Mass

             =36/16=2.25 moles

As 1 mole of methane gives 2 moles of water,

2.25 moles of methane gives=2×2.25=4.5 moles of water.

Amount of H2O produced=4.5×18=81 g