Answer to Question #256199 in General Chemistry for amelia

Question

Gaseous methane CH4 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 15.2 g of methane is mixed with 15. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2

 significant digits.

Answer 1

the balanced equation for the above reaction is;

CH₄ + 2O₂ ---> CO₂ + 2H₂O

Stoichiometry of CH₄ to O₂ is 1:2

The number of methane moles present - 15.2g/ 16 g/mol = 0.95 mol

Number of oxygen moles present - 15 g/ 32 g/mol = 0.47 mol

If methane is the limiting reagent,

0.95 moles of methane react with 0.95x 2 = 1.9 mol 

only 0.47 mol of O₂ is required but 1.9 mol of O₂ has been provided therefore O₂ is in excess and CH₄ is the limiting reactant.

Number of moles of water that can be produced - 0.180 mol

Therefore mass of water produced - 1.9 mol  x 18 g/mol = 34.2 g 

Therefore mass of 34.2 g of water can be produced