Gaseous methane ch4 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o . suppose 0.48 g of methane is mixed with 0.730 g of oxygen. calculate the maximum mass of water that could be produced by the chemical reaction. round your answer to 3 significant digits

1) Balanced chemical equation

CH4 + 2O2 ---> CO2 + 2H2O

2) State molar ratios

1 mol CH4 : 2 mol O2 : 1 mol CO2 : 2 mol H2O

 3) Convert 0.48 g of methane in moles

mole = mass in grams / molar mass

molar mass CH4 = 12g/mol + 4*1g/mol = 16 g/mol

mole = 0.48 g / 16 g/mol = 0.03 mol

4) Convert 0.730 g of oxygen into moles

moles = mass in grams / molar mass

molar mass O2 = 2 * 16g/mol = 32 g/mol

moles = 0.73 g / 32 g/mol = 0.02281 mole

5) Calculate the limiting reactant

Theoretical ratio: 1 mol CH4 : 2 mol O2 = 0.5

Actual ratio: 0.03 mol CH4 : 0.02281 mol O2 = 1.31

=> CH4 is in excess and O2 is the limiting reactant

6) Use the number of moles of the limiting reactant to calculate the number of moles of water produced

2 moles O2 produce 2 moles of H2O, so 0.02281 moles of O2 produce 0.02281 moles of water

7) Convert 0.02281 moles of water into grams

mass in grams = number of moles * molar mass

molar mass of H2O = 2*1g/mol + 16 g/mol = 18 g/mol

mass of H2O = 0.02281 mol * 18 g/mol = 0.4106 g

rounded to 3 significant digits = 0.411 g

0.411 grams

The answer is b i think