How many liters of steam are produced at STP by the combustion of 12.5g of methane

The correct answer is 35 L

Explanation:

The complete combustion of methane (CH₄) gives carbon dioxide (CO₂) and water (H₂O) according to the following chemical equation:

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (g)

The water obtained in gas phase (H₂O (g)) is the steam.

As we can see, 1 mol of CH₄ reacts with 2 moles of O₂ to give 1 mol of CO₂ and 2 moles of H₂O.

We have 12.5 g of CH₄, so we have to first convert the mass to moles of methane by using the molecular weight (Mw) of CH₄

Mw CH₄ = 12 g/mol + (4 x 1 g/mol) = 16 g/mol

moles of CH₄ = mass/Mw = 12.5 g/16 g/mol = 0.78125 moles

According to the chemical equation, the combustion of 1 mol of CH₄ produces 2 moles of H₂O. The water produced is the steam, so we have to multiply by 2 the number of moles we have to obtain the moles of H₂O-steam - we obtain:

0.78125 moles CH₄ x 2 moles H₂O / mol CH₄ = 1.5625 moles H₂O

Finally, we know that 1 mol of gas at STP occupies 22.4 L, so we can determine the liters of steam that are produced as follows:

1.5625 moles H₂O x 22.4 L / 1 mol H₂O = 35 L