limiting reagent (pls show work)

Question

Given the following reaction: (Balance the equation first!)

A) C 3 H 8 + __O 2  __CO 2 + __H 2 O

b) If you start with 14.8 g of C 3 H 8 and 3.44 g of O 2 ,

determine the limiting reagent with respect to CO 2

c) determine the number of moles of CO 2 produced

using 14.8 g of C 3 H 8

d) determine the number of grams of H 2 O produced

using 3.44 g of O 2

e) What is the % yield if you have 1.22 g CO 2 produced

in the lab?

Answer 1

(A) C₃H₈ + 5O₂ ==> 3CO₂ + 4H₂O ... balanced equation

(B) The easiest way to find limiting reagent is to divide moles of reactant by coefficient in balanced eq

moles C3H8 = 14.8 g x 1 mol/44 g = 0.336 moles C3H8 (÷ 1 = 0.336)

moles O2 = 3.44 g O2 x 1 mol O2/32 g = 0.1075 moles O2 (÷ 5 = 0.0215)

So O2 is LIMITING

(C) 0.336 moles C3H8 x 3 moles CO2 / 1 mol C3H8 = 1.00 moles CO2 produced from 14.8 g C3H8

(D) 0.1075 moles O2 x 4 moles H2O / 5 moles O2 = 0.086 moles H2O produced from 3.44 g O2

(E) % yield = actual yield/theoretical yield (x100)

Theoretical yield of CO2 = 0.1075 moles O2 x 3 moles CO2 / 5 moles O2 x 44 g CO2/mol = 2.84 g CO2

% yield = 1.22 g / 2.84 g (x100) = 43.0% yield