Question

calculate the volume occupied by 6.4 grams of methane at stp conditions

Answer 1

Answer:

Explanation:

=6.4/16*22.4=8.96litre

Answer 2

Answer:

Explanation:

Concept:

The concept of organic chemistry will help us to solve the question.

Organic chemistry is the study of the composition, production, reactions, and properties of molecules containing carbon.

Given:

The mass of the methane is $6.4 grams$ and it is in stp condition.

To Find:

The volume of the methane will be found out.

Solution:

First we have to find out the mole of the methane.

Divide the mass of the methane by the molar mass of it to get the number of mole.

$Mole=\frac{Mass}{Molar mass}$

The mass of the methane is  $6.4 grams$ .

The molar mass of the methane is $12 + (4 \times 1) = 16 u$

(Atomic mass of $C$ is $12$ and stomic mass of $H$ is $1$ )

The mole of the methane is $\frac{6.4}{16} =0.4$

Multiply the mole number with molar volume of methane to get the volume of it.

Molar volume of methane at STP is $22.4L$

$V_{CH_{4} } =0.4 \times 22.4 L=8.96L$

The volume occupied by $6.4 grams$ of methane at stp conditions is $8.96L$.

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