Answer to Question #201137 in General Chemistry for Mark

Question

Look for S^o of the reactant and products. Compute for the S^o of the reactant

1.  C₃H₈(g) + 5O₂(g) —› 3CO₂(g) + 4H₂O(g)

Answer 1

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

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

11.2 moles O2 x 3 mol CO2/5 mol O2 x 44 g/mol CO2 = 296 g CO2 theoretical yield

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%