CH4 + 2O2 --> OC2 + 2H2O

What mass of water is formed by burning 4g of methane in excess oxygen?

(Ar of carbon=12, Ar of oxygen=16 and Ar of hydrogen=1)

You transposed some letters; I assume you meant CO2 and not OC2.

mols CH4 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols CH4 to mols H2O.
Then convert mols H2O to grams. grams H2O - mols H2O x molar mass H2O

Post your work if you get stuck.

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.