## HOW TO USE LIMITING REAGENTS

1.Find the limiting reagent

2.Figure out the molar mass of the product

3.compare limiting reagent molar mass with the molar mass of the product an come up with a ratio

4.divide ratio by the limiting reagent number

5.take new ratio and multiply with the limiting reagent MOL and the molar mass of the product

2.Figure out the molar mass of the product

3.compare limiting reagent molar mass with the molar mass of the product an come up with a ratio

4.divide ratio by the limiting reagent number

5.take new ratio and multiply with the limiting reagent MOL and the molar mass of the product

## STEP BY STEP SAMPLE QUESTIONS

Here are some step by step examples! For finding the limiting reactants you have to understand how to solve for the limiting reagent, if you don't know how Click Here!

FIRST QUESTION:

For the balanced equation shown below, if 29.7 grams of C3H8O were reacted with 108 grams of O2, how many grams of H2O would be produced?

2C3H8O+9O2=>6CO2+8H2O

1.Figure out the limiting reagent

2C3H8O + 9O2

60g/mol 32g/mol

.495 mol 3.37 mol

4.455 6.74

2.Figure out the molar mass of H2O

Hx2

Ox1

2+16=18

3.Compare 2C3H8O with 8H2O and find ratio

2:8

4.Divide ratio of 2C3H8O and 8H2O

2 2

1:4

5.Take ratio (1:4) and multiply with the limiting reagent MOL and molar mass of H2O

1:4x.495x18

=35.64

SECOND QUESTION:

For the balanced equation shown below, if 45.9 grams of Al were reacted with 282 grams of HCl, how many grams of AlCl3 would be produced?

2Al+6HCl=>2AlCl3+3H2

1.Figure out the limiting reagent

2Al 6HCl

27g/mol 36.5g/mol

1.7=mol 7.72=mol

10.2 15.44

2.Figure out the molar mass of AlCl3

Alx1

Clx3

27x1

35.5x3

27+106.5=133.5

3.Compare 2Al and 2AlCl3 and find ratio

2 2

4.Divide ratio of 2Al and 2AlCl3

2 2

1:1

2 25.Take ratio (1:1) and multiply with the limiting reagent MOL and molar mass of 2AlCl3

1:1x1.7x133.5

=226.95

FIRST QUESTION:

For the balanced equation shown below, if 29.7 grams of C3H8O were reacted with 108 grams of O2, how many grams of H2O would be produced?

2C3H8O+9O2=>6CO2+8H2O

1.Figure out the limiting reagent

2C3H8O + 9O2

__29.7g____108g__60g/mol 32g/mol

.495 mol 3.37 mol

__x 9____x 2__4.455 6.74

**The limiting reagent is 2C3H8O**2.Figure out the molar mass of H2O

Hx2

Ox1

2+16=18

3.Compare 2C3H8O with 8H2O and find ratio

**2**C3H8O**8**H2O2:8

4.Divide ratio of 2C3H8O and 8H2O

__2__:__8__2 2

1:4

5.Take ratio (1:4) and multiply with the limiting reagent MOL and molar mass of H2O

1:4x.495x18

=35.64

SECOND QUESTION:

For the balanced equation shown below, if 45.9 grams of Al were reacted with 282 grams of HCl, how many grams of AlCl3 would be produced?

2Al+6HCl=>2AlCl3+3H2

1.Figure out the limiting reagent

2Al 6HCl

__45.9g____282g__27g/mol 36.5g/mol

1.7=mol 7.72=mol

__x 6____x 2__10.2 15.44

**The limiting reagent is 2Al**2.Figure out the molar mass of AlCl3

Alx1

Clx3

27x1

35.5x3

27+106.5=133.5

3.Compare 2Al and 2AlCl3 and find ratio

**2**Al**2**AlCl3__2__:__2__2 2

4.Divide ratio of 2Al and 2AlCl3

__2__:__2__2 2

1:1

2 25.Take ratio (1:1) and multiply with the limiting reagent MOL and molar mass of 2AlCl3

1:1x1.7x133.5

=226.95