To determine the mass of water formed by burning 4g of methane (CH4), we'll need to calculate the stoichiometry of the reaction.
Step 1: Calculate the molar mass of CH4.
Molar mass of CH4 = (1 x Ar of C) + (4 x Ar of H)
= (1 x 12) + (4 x 1)
= 12 + 4
= 16 g/mol
Step 2: Calculate the moles of CH4.
Moles of CH4 = Mass of CH4 / Molar mass of CH4
= 4g / 16g/mol
= 0.25 mol
Step 3: Use the stoichiometric ratio to determine the moles of water formed.
From the balanced equation:
1 mole of CH4 produces 2 moles of H2O
0.25 mol of CH4 will produce (0.25 x 2) mol of H2O
= 0.5 mol of H2O
Step 4: Calculate the mass of water formed.
Molar mass of H2O = (2 x Ar of H) + (1 x Ar of O)
= (2 x 1) + (1 x 16)
= 2 + 16
= 18 g/mol
Mass of H2O = Moles of H2O x Molar mass of H2O
= 0.5 mol x 18 g/mol
= 9 g
Therefore, the mass of water formed by burning 4g of methane in excess oxygen is 9g.