Answer to Question #248190 in General Chemistry for Potterhead

Question

Gaseous methane CH4  will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 6.3 g of methane is mixed with 15.0 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.

Answer 1

CH₄ + 2O₂ → CO₂ + 2H₂O

M(CH₄) = 16.04 g/mol

n(CH₄) $= \frac{6.3}{16.04}=0.393 \;mol$

M(O₂) = 15.99 g/mol

n(O₂) $= \frac{15.0}{15.99}=0.938 \;mol$

For each mole of CH₄ we need only 2 moles of O₂, but for 0.393 mol of CH₄ we have more O₂ than needed. So, CH₄ is the limiting reactant.

According to the reaction:

n(H₂O) = 2n(CH₄) $= 2 \times 0.393 = 0.786 \;mol$

M(H₂O) = 18.01 g/mol

m(H₂O) $= 18.01 \times 0.786 = 14.15 \;g$

Answer: 14.1 g.