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

Question

Equation: C3H8 + 5O2 = 3CO2 + 4H2O (already balanced)

 

a) If you start with 14.8 g of C3H8 and 3.44 g of O2, determine the limiting reagant

 

b) Determine the number of moles of carbon dioxide produced

 

c) Determine the number of grams of H2O produced

 

d) Determine the number of grams of excess reagant left

Answer 1

C3H8 + 5O2 = 3CO2 + 4H2O

a)  moles C3H8 present = 14.8 g x 1 mol/44 g = 0.336 moles

moles O2 present = 3.44 g x 1 mole/32 g = 0.1075

From balanced equation you need 5 x as much O2 as you have C3H8.  Do you?  NO.  You have less.

Limiting reactant is O2

 

b)  moles of CO2 produced will now be dependent (limited) by the amount of O2 present

moles CO2 = 0.1075 moles O2 x 3 moles CO2/5 moles O2 = 0.0645 moles CO2 produced

 

c)  moles H2O produced = 0.1075 moles O2 x 4 moles H2O/5 moles O2 = 0.086 moles H2O

mass of H2O = 0.086 moles x 18 g/mole = 1.55 g H2O

 

d)  From balanced equation you use 1 mole C3H8 for every 5 moles of O2

moles of C3H8 used = 0.1075 moles O2 x 1 mole C3H8/5 moles O2 = 0.0215 moles C3H8 used

moles C3H8 remaining = 0.336 - 0.0215 = 0.3145 moles 

mass C3H8 remaining = 0.3145 moles x 44 g/mole = 13.8 g (the reaction used only 1 g of C3H8)