## WHAT EXACTLY IS A LIMITING REAGENT?

The limiting reagent is that reactant that runs out first and determines the end of a reaction.

## LIMITING REAGENTS IN EVERYDAY LIFE

When it is cold outside you want to wear some warm mittens to keep your hands cold. If you bought 10 pairs of mittens and have 14 people the mittens would be the limiting reagent because there isn't enough gloves for each person.

## HOW TO FIND THE LIMITING REAGENT

1.First write down the unbalanced equation

2. Write the grams for each element which is given

3.Next find the GFW of each element. If the element has a coefficient in the front forget about that for now.

4.Once you find the GFW of each element, divide the GFW by that elements grams to get each elements total MOL

5.Now that you have the total MOL take the coefficients of each element and multiply it to the other element total MOL

6.Take the totals of each element and compare. The one with a smaller amount is the limiting reagent

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2. Write the grams for each element which is given

3.Next find the GFW of each element. If the element has a coefficient in the front forget about that for now.

4.Once you find the GFW of each element, divide the GFW by that elements grams to get each elements total MOL

5.Now that you have the total MOL take the coefficients of each element and multiply it to the other element total MOL

6.Take the totals of each element and compare. The one with a smaller amount is the limiting reagent

Feel like you understand, Click Here to practice some equations!!!

## STEP BY STEP EXAMPLE QUESTIONS

Here is some step by step example questions!

FIRST QUESTION:

For the balanced equation shown below, what would be the limiting reagent if 24.0 grams of C2H4O were reacted with 27.0 grams of O2?

2C2H4O+5O2=>4CO2+4H2OC2H4OO2

___C2H4O ___ O2

1.First write down the unbalanced equation

2C2H4O+5O2

2. Write the grams for each element which is given

2C2H4O + 5O2

24.0g 27.0g

3.Next find the GFW of C2H4O and O2

C2H4O

carbon-16x2

hydrogen-1x4

oxygen-16x1

32+4+16=44

oxygen-16x2

32

4.Once you find the GFW of each element, divide it by the grams

2C2H4O + 5O2

44 g/mol 32 g/mol

.545MOL .843MOL

5.Now take the coefficients of each element (For 2C2H4O its 2, for 5O2 its 5) and multiply it to the other element total MOL

.545 .843

2.725 1.686

6.Take the totals of each element and compare. The one with a smaller amount is the limiting reagent

2C2H4O +

2.725

FIRST QUESTION:

For the balanced equation shown below, what would be the limiting reagent if 24.0 grams of C2H4O were reacted with 27.0 grams of O2?

2C2H4O+5O2=>4CO2+4H2OC2H4OO2

___C2H4O ___ O2

1.First write down the unbalanced equation

2C2H4O+5O2

2. Write the grams for each element which is given

2C2H4O + 5O2

24.0g 27.0g

3.Next find the GFW of C2H4O and O2

C2H4O

carbon-16x2

hydrogen-1x4

oxygen-16x1

32+4+16=44

oxygen-16x2

32

4.Once you find the GFW of each element, divide it by the grams

2C2H4O + 5O2

__24.0g____27.0g__44 g/mol 32 g/mol

.545MOL .843MOL

5.Now take the coefficients of each element (For 2C2H4O its 2, for 5O2 its 5) and multiply it to the other element total MOL

.545 .843

__x 5____x 2__2.725 1.686

6.Take the totals of each element and compare. The one with a smaller amount is the limiting reagent

2C2H4O +

**5O2**2.725

**1.686**SECOND QUESTION:

For the balanced equation shown below, what would be the limiting reagent if 80.1 grams of BiCl3 were reacted with 18.4 grams of H2S?

2BiCl3+3H2S=>Bi2S3+6HCl

1.First write down the unbalanced equation

2BiCl3+3H2S

__BiCl3 __H2S

2. Write the grams for each element which is given

2BiCl3 + 3H2S

80.1g 18.4

3.Next find the GFW of BiCl3 and H2S

BiCl3

Bismuth-209x1

chloride-35 x 3

209+105=314

H2S

hydrogen-1x2

Sulfur-32x1

2+32=34

4.Once you find the GFW of each element, divide it by the grams

2BiCl3 + 3H2S

314g/mol 34g/mol

.255MOL .541MOL

5.Now take the coefficients of each element (For 2BiCl3 its 2 and for 3H2S its 3) and multiply it to the other element total MOL

.255 .541

.765 1.082

6.Take the totals of each element and compare. The one with a smaller amount is the limiting reagent

Once you feel like you have a good understanding, Click Here to do some practice problems

For the balanced equation shown below, what would be the limiting reagent if 80.1 grams of BiCl3 were reacted with 18.4 grams of H2S?

2BiCl3+3H2S=>Bi2S3+6HCl

1.First write down the unbalanced equation

2BiCl3+3H2S

__BiCl3 __H2S

2. Write the grams for each element which is given

2BiCl3 + 3H2S

80.1g 18.4

3.Next find the GFW of BiCl3 and H2S

BiCl3

Bismuth-209x1

chloride-35 x 3

209+105=314

H2S

hydrogen-1x2

Sulfur-32x1

2+32=34

4.Once you find the GFW of each element, divide it by the grams

2BiCl3 + 3H2S

__80.1g____18.4__314g/mol 34g/mol

.255MOL .541MOL

5.Now take the coefficients of each element (For 2BiCl3 its 2 and for 3H2S its 3) and multiply it to the other element total MOL

.255 .541

__x 3____x 2__.765 1.082

6.Take the totals of each element and compare. The one with a smaller amount is the limiting reagent

**2BiCl3**+ 3H2S**.765**1.082Once you feel like you have a good understanding, Click Here to do some practice problems