Question

How many moles of O2 (molar mass = 32.00 g/mol) are needed to react completely with 52.0 L of CH4 (g) at STP to produce CO2 and H2O? (molar mass of CH4 16.04 g/mol)
a. 11.6
b. 2.32
c. 4.64
d. 52.0

Answer 1

Answer: The moles of oxygen gas needed to react are 4.64 moles.

Explanation:

We are given:

Volume of methane gas = 52.0 L

STP conditions:

22.4 L of volume is occupied by 1 mole of a gas

So, 52.0 L of volume will be occupied by = $\frac{1}{22.4}\times 52.0=2.32mol$ of methane gas

The chemical equation for the combustion of methane follows:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

By Stoichiometry of the reaction:

1 mole of methane reacts with 2 moles of oxygen gas

So, 2.32 moles of methane will react with = $\frac{2}{1}\times 2.32=4.64mol$ of oxygen gas

Hence, the moles of oxygen gas needed to react are 4.64 moles.

Answer 2

CH4 (g) + 2O2 (g) --> CO2 (g) + 2H2O (i)
at STP, 1 mol gas = 22.4 L gas. so you have 52.0 L / 22.4 L / mol = 2.32 mol CH4.
According to  the balanced reaction, you need 2 mol O2 for every 1 mol CH4. so you need 2 x 2.32 mol = 4.64 mol O2.

hope this help