During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.

What is the amount of the leftover reactant?

0.74 grams of methane

0.89 grams of methane

1.22 grams of oxygen

1.45 grams of oxygen

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.

it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of CH₄ needs → 2 mol of O₂

? mol of CH₄ needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 - 0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over * molar mass

= 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane