Question

Calculate the amount of oxygen(g) required for the combustion of 32g of methane

Please reply fast​

Answer 1

$\huge \text{\underline {ANSWER}}$

Amount of oxygen needed for combustion of 32g methane = 128g

$\huge \text{\underline {EXPLANATION}}$

The combustion of methane is given by the equation,

CH₄ + 2O₂ -------> CO₂ + 2H₂O

From the equation we understand that,

1 mole CH₄ requires 2 mole O₂

Also,

Given mass of CH₄ = 32g

Molar mass of CH₄ = 16g

Hence

Number of moles of CH₄ = Given mass / Molar mass

Number of moles of CH₄ = 32 / 16

Number of moles of CH₄ = 2

Since,

1 mole CH₄ requires 2 mole O₂

2 mole CH₄ requires 2*2 = 4 mole O₂

Number of moles of O₂ = Given mass / Molar mass

4 = Given mass / 32

Given mass = 4*32

Given mass = 128g