Writing And Balancing Chemical Reactions, Stoichiometry, Limiting Reactants Worksheet With Answers - Chem110 Page 6
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8CHEM110 Week 5 Notes (Stoichiometry)
Page 6 of 8
When 155g of ammonium nitrate decomposes, what is the theoretical yield of dinitrogen
oxide and water?
Step 1: Write the balanced chemical equation
Step 2: Convert the mass of reactant to moles
Step 3: Convert moles of reactant to moles of product
Step 4: Convert moles of product to grams to get the theoretical yield
Step 1:
NH
NO
→ N
O + 2H
O
4
3
2
2
mass, g
155
____ ____
mol
___
____ ____
Steps 2, 3, and 4 for N
O
2
155 g NH
NO
1 mol NH
NO
1 mol N
O
44.013 g N
O
4
3
4
3
2
2
80.043 g NH
NO
1 mol NH
NO
1 mol N
O
4
3
4
3
2
= 85.2g N
O (3SD) is the theoretical yield.
2
Steps 2, 3, and 4 for H
O
2
155 g NH
NO
1 mol NH
NO
2 mol H
O
18.015 g H
O
4
3
4
3
2
2
80.043 g NH
NO
1 mol NH
NO
1 mol H
O
4
3
4
3
2
= 67.8g H
O (3SD) is the theoretical yield. You can also use the Law of Conservation of
2
Mass
The answers from each step appear in the table below.
NH
NO
→ N
O + 2H
O
4
3
2
2
mass, g
155
85.2 69.8
mol
1.936
1.936 3.873
In CHEM II, we will use before, change, after (BCA) tables to track reactants and
products. A BCA table for the example above appears below.
NH
NO
→ N
O + 2H
O
4
3
2
2
Before, g
155
0
0
Before, mol
1.936
0
0
Change, mol -1.936
+1.936 +3.873
After, mol
0
1.936 3.873
After, g
0
85.2
69.8
If 74.2g of N
O were obtained by the previous experiment, what was the percent yield?
2
To calculate percent yield, divide the actual yield from the experiment by the theoretical
yield. %yield = 74.2/85.2 * 100% = 87.1% yield.
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