Stoichiometry And Percent Yield of C3H8+ 5O2 = 3Co2 + 4H2O Please answer 1-5 (work would be nice) Thank you

Question

C3H8+ 5O2 = 3Co2 + 4H2O

1. Calculate the expected theoretical yield of Co2 in Moles

A.If 1.62 Moles of Of C3H5 were used

B. If 11.6 Moles of O2 were used

2.As in the above reactions, if both 1.62 Moles of C3H8 and 11.6 Moles of O2 were used in a reaction, what is the limiting reaction. Justify your answer

3.Calculate the percent yield of 209 Grams of Co2 were actually produced.

Answer 1

You are given the balanced equation, so no need to balance it.

C3H8 + 5O2 ===> 3CO2 + 4H2O

1A). 1.62 moles C3H5 x 3 moles CO2/1 mole C3H8 x 44g/mol CO2 = 214 g CO2 theoretical yield

1B). 11.2 moles O2 x 3 mol CO2/5 mol O2 x 44 g/mol CO2 = 296 g CO2 theoretical yield

2). Limiting reactant would be C3H5 because it produces the least amount of product, so it will run out first.

3.) Percent yield = actual yield/theoretical yield (x100%) = 209 g/214 g (x100%) = 97.7%