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
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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
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
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
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 1.082
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 1.082
Once you feel like you have a good understanding, Click Here to do some practice problems