Question

Gaseous methane CH4 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 10.2g of water is produced from the reaction of 13.8g of methane and 30.3g of oxygen gas, calculate the percent yield of water.

Answer 1

32.9%

Step-by-step explanation:

The balanced chemical equation for the reaction is:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

From the balanced equation, we can see that 1 mole of methane (CH4) reacts with 2 moles of oxygen (O2) to produce 1 mole of carbon dioxide (CO2) and 2 moles of water (H2O).

First, let’s calculate the theoretical yield of water. The molar mass of methane (CH4) is approximately 16.04 g/mol. So, 13.8 g of methane is approximately 0.86 moles.

Since the reaction produces 2 moles of water for every mole of methane, the theoretical yield is 0.86 * 2 = 1.72 moles of water.

The molar mass of water (H2O) is approximately 18.02 g/mol. So, the theoretical yield in grams is 1.72 moles * 18.02 g/mol = approximately 31 g.

The percent yield is calculated as follows:

Percent Yield = (Actual Yield / Theoretical Yield) * 100%

Substituting the given values:

Percent Yield = (10.2 g / 31 g) * 100% = approximately 32.9%