The reaction between 32.0 g of CH4 and excess oxygen gas produces 75.9 g of CO2 gas and some water. Determine the percent yield

Question

 How many moles of CH, reacted? mass CH₁ = 32.0g/16.0'lg/mol Moles C*n_{n} = 2 moles

b How many moles of CO2 could theoretically be produced 1mol and 2mol for I mol O2 H₂0. For 2.00 Moles CH4, we'll have 2.00moles

c. What is the actual yield of CO2?

d. What is the percent yield?

e. What is the theoretical yield of CO2?

So the thing is the last three I don't know how to solve because I just learn this yesterday and I'm expected to do a test on this

Answer 1

Write the balanced equation for the reaction taking place:

CH₄ + 2O₂ ==> CO₂ + 2H₂O ... balanced equation

b) Theoretical yield of CO2:

32.0 g CH4 x 1 mol CH4/16 g = 2 mols CH4 present

2 mol CH4 x 1 mol CO2 / 1 mol CH4 = 2 mols CO2 can theoretically be produced

c) The actual yield of CO2:

75.9 g CO2 is the actual yield in grams, as given in the problem

75.9 g CO2 x 1 mol CO2 / 44 g = 1.73 moles CO2 is the actual yield in moles

d) Percent yield is actual yield / theoretical yield (x100%)

1.73 mols / 2.00 mols (x100%) = 86.5% yield

e) Theoretical yield of CO2 is given in (b) above. It is 2 mols CO2