Question

(a)calculate the theoretical yield of carbon dioxide from the reaction of 33.3 g of propane (C3H8) and 25.0 g of oxygen gas. (write balanced chemical equation).

(b) How many grams of excess reagent would be left over?

Answer 1

a)

the reaction

C3H8 + O2 = CO2 + H2O

balance C

C3H8 + O2 = 3CO2 + H2O

balacne H

C3H8 + O2 = 3CO2 + 4H2O

blance O

C3H8 + 5O2 = 3CO2 + 4H2O

so..

MW = 44.1 g/mol

mol of propane = mass/MW =33.3/44.1 = 0.75510 mol of propane

mol of O2 = mass/M W= 25/32 = 0.78125

ratio is 1:5 so O2 is limiting

0.78125 mol of O2 --> 3/5*0.78125 = 0.46875 mol of CO2 will form

mass = mol*MW = 0.46875*44 = 20.625 g of CO2

b)

grams of excess reactant:

0.75510 - 0.78125/5 = 0.59885 mol of propane left

mass = mol*MW = 0.59885*44 = 26.3494 g