imiting Reagents (Chemistry) Answer a-d + Show Work Please

Question

Given the balanced equation: C3H8 + 5O2 ----> 3CO2 + 4H2O

a) When 45g C3H8 react with 65g O2, how many grams of CO2 can be produced?

b) What is the limiting agent?

c) How many grams of the excess reactant remain?

d) What is the percent yield of this reaction if 45.0 grams of Cl2 are obtained in lab?

Answer 1

C₃H₈ + 5O₂ ----> 3CO₂ + 4H₂O

Before you can answer (a), you must first answer (b)

(b) Limiting reactant:

45 g C₃H₈ x 1 mol/44 g = 1.02 moles

65 g O₂ x 1 mol/32 g = 2.03 moles but it takes 5 moles O₂ for each 1 mol of C₃H₈, so not enough O₂.

O₂ is LIMITING

(a) Grams CO₂ produced:

2.03 moles O₂ x 3 moles CO₂ / 5 moles O₂ x 44 g CO₂/mol = 53.6 g CO₂

(c) Excess reagent is C3H8:

Started with 1.02 moles

moles used up = 2.03 mol O2 x 1 mol C3H8 used/5 mol O2 = 0.406 moles used

moles left over = 1.02 mol - 0.406 mol = 0.614 moles left

grams left = 0.614 moles x 44 g/mol = 27.0 g remaining

(d) Percent yield = actual yield/theoretical yield (x100%) but you probably meant to write CO₂, not Cl₂

% yield = 45.0 g/53.6 g (x100%) = 83.95% = 84.0% yield