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skad [1K]
3 years ago
13

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
Chemistry
2 answers:
Gre4nikov [31]3 years ago
8 0

<u>Answer:</u> The moles of oxygen gas needed to react are 4.64 moles.

<u>Explanation:</u>

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.

Rainbow [258]3 years ago
6 0
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
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