what mass of water is produced from the complete combustion of 6.20x10-3g of methane

Answer 1

To solve this problem, we first need to write the balanced chemical equation for the combustion of methane:

CH4 + 2O2 -> CO2 + 2H2O

From the equation, we can see that 1 mole of methane produces 2 moles of water.

First, we need to calculate the number of moles of methane in 6.20x10^-3g:

Molar mass of CH4 = 12.01 + 1.01 x 4 = 16.05 g/mol

Number of moles of CH4 = 6.20x10^-3g / 16.05 g/mol = 3.86x10^-4 mol

Since 1 mole of methane produces 2 moles of water, the number of moles of water produced will be twice that of methane:

Number of moles of H2O = 3.86x10^-4 mol x 2 = 7.72x10^-4 mol

Now, we can calculate the mass of water produced:

Molar mass of H2O = 18.02 g/mol

Mass of water produced = 7.72x10^-4 mol x 18.02 g/mol = 0.0139g

Therefore, 0.0139g of water is produced from the complete combustion of 6.20x10^-3g of methane.