Limiting Reagent - When Reactants are mixed together in the wrong ratio one will be in excess and the other will be limiting.

What volume of CO2 is produced when 190g of C3H8 (44g/mol) is reacted with 400L of O2 at STP?

Moles: 4.32 17.86 10.72

C3H8 + 5 O2 3 (CO2) + 4 (H20)

Other: 190g 400L 240L

4.32 moles of C3H8 = x moles of CO2

4.32 * 3 = 12.96 Excess Reagent

17.86 moles of O2 = 3/5 CO2

17.86 * 3/5= 10.72 Limiting Reagent

ALL STOICHIOMETRY IS BASED UPON THE LIMITING REAGENT

Moles: 4.32 17.86 10.72

C3H8 + 5 02 3 (CO2) + 4 (H20)

Other: 190g 400L 240L Theoretical yield

A student carried out the above experiment and collected 198L

of CO2. What is the student’s percent yield?

Percent yield= (actual results/theoretical yield) *100

(198/240) * 100 = 82.5% percent yield (also finds precision)

What mass of C3H8 is unreacted?

Take the moles of the limiting reagent (17.86) and divide it by

the molar ratio of the excess to limiting (1/5mol).

17.86/5= 3.57 moles of C3H8 Used

Then subtract the moles used from the moles given

4.32 – 3.57 = .75

Multiply the difference by the gfm (gram formula mass)

X=.75 * 44 g/mol

X = 33g excess

HW Question 34 + this question

2 (NaCl) + F2 2 (NaF) + Cl2

1) How many grams of Cl2 (71 g/mol) are produced when 60g

of NaCl + 60 g of F2 are reacted.

2) A student recovers 30.5 g of Cl2, what is the percent yield?

3) What is the mass of the excess reactant remains?