what mass of oxygen is needed for the complete combustion of 6.20x10-3g of methane
First, we need to determine the balanced chemical equation for the combustion of methane (CH4) to calculate the amount of oxygen required.
The balanced chemical equation for the combustion of methane is:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we can see that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.
1 mole of CH4 has a mass of 16 g, so 6.20 x 10^-3 g of CH4 is:
6.20 x 10^-3 g /16 g/mol = 3.875 x 10^-4 moles of CH4
To calculate the amount of oxygen needed for the complete combustion of CH4, we use the mole ratio from the balanced chemical equation. Since the mole ratio of CH4 to O2 is 1:2, then:
3.875 x 10^-4 moles of CH4 x 2 moles of O2/ 1 mole of CH4 = 7.75 x 10^-4 moles of O2
To determine the mass of oxygen needed, we multiply the number of moles of O2 by the molar mass of O2 (32 g/mol):
7.75 x 10^-4 moles x 32 g/mol = 2.48 x 10^-2 g
Therefore, 2.48 x 10^-2 g of oxygen is needed for the complete combustion of 6.20 x 10^-3 g of methane.