What mass of oxygen is needed for the complete combustion of 8.90×10−3 of methane?

What was not understood by my previous response?

To determine the mass of oxygen needed for the complete combustion of methane, we need to balance the chemical equation for the combustion reaction:

CH4 + 2O2 -> CO2 + 2H2O

From the balanced equation, we can see that one molecule of methane (CH4) reacts with two molecules of oxygen (O2) to produce one molecule of carbon dioxide (CO2) and two molecules of water (H2O).

The molar mass of methane (CH4) is 16.04 grams/mol, and the molar mass of oxygen (O2) is 32.00 grams/mol.

To find the mass of oxygen needed, we need to convert the given amount of methane (8.90×10−3 mol) to moles of oxygen using the mole ratio from the balanced equation.

Given:
Amount of methane (CH4) = 8.90 × 10−3 mol

From the balanced equation:
1 mol CH4 reacts with 2 mol O2

To find the moles of oxygen (O2):
Moles of O2 = (8.90 × 10−3 mol CH4) * (2 mol O2 / 1 mol CH4) = 1.78 × 10−2 mol O2

Finally, to determine the mass of oxygen:
Mass of O2 = Moles of O2 × Molar mass of O2
= (1.78 × 10−2 mol) × (32.00 g/mol)
= 0.5696 g

Therefore, the mass of oxygen needed for the complete combustion of 8.90 × 10−3 mol of methane is approximately 0.5696 grams.