CH4 (g) + O2 (g) → CO2 (g) + H2O (g) to. If 28.6 g of CH4 is reacted with 57.6 g of O2, calculate the number of grams of CO2 produced. b. If you actually get 32.1 g of CO2, calculate the percent yield. c. Calculate the number of moles of excess reagent remaining at the end of the reaction.

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Explanation:

First we must obtain the limiting reactant. The equation of the reaction is;

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

Number of moles of CH4= 28.6 g/16g/mol = 1.8 moles

Since the reaction is 1:1, 1.8 moles of CO2 was produced

Number of moles of O2 = 57.6 g/32 g/mol = 1.8 moles

2 moles of O2 produced 1 mole of CO2

1.8 moles of O2 produced 1.8 × 1/2 = 0.9 moles of CO2

Hence O2 is the limiting reactant

Mass of CO2 produced = 0.9 moles × 44 g/mol = 39.6 g

%yield = 32.1g/39.6 g × 100

%yield = 81.1%

According to the reaction equation;

2moles of O2 reacts with 1 mole of CH4

1.8 moles of O2 reacts with 1.8 × 1/2 =0.9 moles of CH4

Number of moles of CH4 left = 1.8 moles - 0.9 moles

Number of moles of CH4 left = 0.9 moles

c. is the answer

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