What mass of water is produced from the complete combustion of 7.50×10−3 g of methane?
To determine the mass of water produced from the complete combustion of methane, we first need to write and balance the equation for the combustion of methane:
CH4 + 2O2 -> CO2 + 2H2O
From the balanced equation, we see that 1 mole of methane produces 2 moles of water.
First, calculate the number of moles of methane:
1 mole of CH4 has a mass of 16.04 g
7.50 x 10^(-3) g of CH4 is equivalent to:
7.50 x 10^(-3) g / 16.04 g/mol = 4.68 x 10^(-4) mol of CH4
Using the mole ratio from the balanced equation, we can determine the number of moles of water produced:
4.68 x 10^(-4) mol CH4 x (2 mol H2O / 1 mol CH4) = 9.36 x 10^(-4) mol H2O
Now, calculate the mass of water produced:
1 mole of H2O has a mass of 18.02 g
9.36 x 10^(-4) mol of H2O is equivalent to:
9.36 x 10^(-4) mol x 18.02 g/mol = 1.68 x 10^(-2) g of H2O
Therefore, the mass of water produced from the complete combustion of 7.50 x 10^(-3) g of methane is 1.68 x 10^(-2) g.