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

Evaporation: this is when warmth from the sun causes water from oceans, lakes, streams, ice and soils to rise into the air and turn into water vapour (gas). water vapour droplets join together to make clouds! condensation: this is when water vapour in the air cools down and turns back into liquid water.