If the percent yield for the following reaction is 65.0%, how many grams of KClO3 are needed to produce 32.0 g of O2?
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
Answers
Explanation:
165 g KClO3
Explanation:
Once again, start with the balanced chemical equation for this decomposition reaction
2KClO3(s] →2KCl(s] +3O(g]
Notice that you have a 2:3
mole ratio between potassium chlorate,
KClO3, and oxygen gas,
O2
.
This means that for a reaction that has an 100% percent yield, every two moles of potassium chlorate will produce three moles of oxygen gas.
Keep this in mind.
So, you know that your reaction must produce
42.0 g
of oxygen gas. Use oxygen gas' molar mass to find how many moles must be produced 42.0g⋅
1 mole O2
32.0
g
=
1.3125 moles O
2
So, how many moles of potassium chlorate would you need If the reaction had an 100% percent yield?
Use the aforementioned mole ratio to find
1.3125
moles O
2
⋅
2
moles KClO3
3
moles O
2
=
0.875 moles KClO
3
However, you know for a fact that the percent yield of the reaction is not 100%, but 65.0%. This means that you will need to use more potassium chlorate to produce this much oxygen gas.
Percent yield is defined as the actual yield of the reaction divided by the theoretical yield of the reaction.
% yield
=
actual yield
theoretical yield
×
100
You know that the reaction's actual yield is 42.0 g of oxygen gas, which means that the theoretical yield must be
65.0%
=
42.0 g
theoretical yield
×
100
theoretical yield
=
42.0 g
⋅
100
65
=
64.6 g
This means that you need to find how many grams of potassium chlorate would theoretically produce 64.6 g of oxygen gas.
Once again, use oxygen's molar mass and the mole ratio that exists between the two compounds
64.6
g O
2
⋅
1 mole O
2
32.0
g O
2
=
2.019 moles O
2
This means that you need to use
2.019
moles O
2
⋅
2
moles KClO
3
3
moles O
2
=
1.346 moles KClO
3
moles of potassium chlorate. Finally, use the compound's molar mass to find how many grams would contain this many moles
1.346
moles KClO
3
⋅
122.55 g
1
mole KClO
3
=
165 g KClO
3