If 128 grams of CH4 and 160 grams of O2 react, how many moles of CO2 will form. What would be the limiting reactant/excess reactant and theoretical yield?

Answer 1

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1. Step 1. Write out the balanced chemical equation

CH₄ + 2O₂ → CO₂ + 2H₂O

1. Step 2. Find the limiting reagent

moles of CO₂ from CH₄ = 128 g of CH₄ * 1 mol of CH₄/ 16.04 g of CH₄ * 1 mol of CO₂/ 1 mol of CH₄ =

= 7.98 mol of CO₂ from CH₄

moles of CO₂ from O₂ = 160 g of O₂ * 1 mol of O₂/ 32.00 g of O₂ * 1 mol of CO₂/ 2 mol of O₂ =

= 2.50 mol of CO₂ from O₂ (O₂ is limiting reagent)

1. Step 3. Find mass of CO₂

mass of CO₂ = 2.50 mol of CO₂ * 44.01 g CO₂/1 mol of CO₂ = 110 g of CO₂

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Answer 2

One way to do this is to calculate the moles of CO₂ from each reactant as if it is the LR. Whichever one gives the smaller value of CO₂ produced is the limiting reactant and you have calculated the amount of CO₂ produced.

Since the reaction is complete we actually can work out the stoichiometry from the chemical formulas

128 grams CH₄ (1mole/16.0 g CH₄) (1 mole C/mole CH₄)(1mole CO₂/mole C) = 8 moles CO₂

160 grams O (1mole/32 g O₂) (1mole CO₂/mole O₂) = 5 moles CO₂ (O₂ is LR and 5 moles is the answer)

The last part could have been 2 moles O/ 1 mole O₂)(1 mole CO₂/2 moles O) which is the same.