<h3>
Answer:</h3>
No masses of CH₄ and O₂ remained after the reaction, while 22.005 g of CO₂ and 18.02 g of H₂O remained
<h3>
Explanation:</h3>
The combustion of methane is given by the reaction;
CH₄(g)+2O₂(g) → CO₂(g)+2H₂O(g)
We are given, 8 g of CH₄ and 16.00 g of O₂
Required to determine the mass of CH₄, O₂, CO₂ and H₂O that remained after the complete reaction.
<h3>Step 1: Moles of CH₄ and O₂ in the mass given </h3>
Moles = mass ÷ molar mass
Molar mass of CH₄ = 16.04 g/mol
Moles of CH₄ = 8.00 g ÷ 16.04 g/mol
= 0.498 moles
= 0.5 moles
Molar mass of O₂ = 16.0 g/mol
Moles of O₂ = 16.00 g ÷ 16.00 g/mol
= 1 mole
From the reaction, 1 mole of CH₄ reacts with 2 moles of O₂
CH₄ is the limiting reactant since it is way less than the amount of O₂
Therefore, 0.5 moles of CH₄ will react with 1 mole of oxygen.
This means there will be no amount of O₂ and CH₄ that remains.
<h3>Step 2: Moles of CO₂ and H₂O that were produced.</h3>
From the reaction 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O.
Therefore,
In our case, 0.5 moles of CH₄ will react with 1 mole of O₂ to produce 0.5 moles of CO₂ and 1 mole of H₂O.
<h3>Step 3: Mass of CO₂ and H₂O produced </h3>
Mass = Moles × Molar mass
Molar mass of CO₂ = 44.01 g/mol
Mass of CO₂ = 0.5 mol × 44.01 g/mol
= 22.005 g
Molar mass of H₂O = 18.02 g/mol
Moles of H₂O = 1 mole × 18.02 g/mol
= 18.02 g
Therefore, no masses of CH₄ and O₂ remained after the reaction, 22.005 g of CO₂ and 18.02 g of H₂O remained
