what mass of oxygen is needed for the complete combustion of 7.50x10-3g of methane
To calculate the mass of oxygen needed for the complete combustion of methane, we first need to write and balance the chemical equation for the combustion of methane (CH4).
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that 1 mole of methane (CH4) requires 2 moles of oxygen (O2) for complete combustion.
First, calculate the molar mass of CH4:
C = 12.01 g/mol
H = 1.01 g/mol
4(H) = 4.04 g/mol
Molar mass of CH4 = 12.01 + 4.04 = 16.05 g/mol
Next, calculate the number of moles of methane in 7.50x10-3g:
number of moles = mass / molar mass
number of moles = 7.50x10-3g / 16.05 g/mol
number of moles = 4.67x10-4 moles
Since 1 mole of CH4 requires 2 moles of O2, the number of moles of O2 needed for the combustion is:
number of moles of O2 = 2 * number of moles of CH4
number of moles of O2 = 2 * 4.67x10-4 moles
number of moles of O2 = 9.34x10-4 moles
Finally, calculate the mass of oxygen needed:
mass of O2 = number of moles of O2 * molar mass of O2
mass of O2 = 9.34x10-4 moles * 32 g/mol
mass of O2 = 0.0299 g
Therefore, the mass of oxygen needed for the complete combustion of 7.50x10-3g of methane is 0.0299g.