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Steps in Doing a Stoichiometry Problem By: Devin Campbell. How many total grams of Copper II Nitride and Oxygen are needed to produce 3.27 grams of Copper II Oxide?. 1. Write a complete and balanced chemical equation. 2Cu 3 N 2 + 3O 2 -----> 6CuO + 2N 2. 2. Draw a column after each chemical. - PowerPoint PPT Presentation
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Steps in Doing a Stoichiometry
Problem
By: Devin Campbell
Steps in Doing a Stoichiometry
Problem
By: Devin Campbell
How many total grams of Copper II Nitride and Oxygen are needed to
produce 3.27 grams of Copper II Oxide?
How many total grams of Copper II Nitride and Oxygen are needed to
produce 3.27 grams of Copper II Oxide?
1. Write a complete and balanced chemical
equation
1. Write a complete and balanced chemical
equation
• 2Cu3N2 + 3O2 -----> 6CuO + 2N2• 2Cu3N2 + 3O2 -----> 6CuO + 2N2
2. Draw a column after each chemical
2. Draw a column after each chemical
2Cu3N2 + 3O2---> 6CuO +
2N2
3. Write the amount given in the correct column
3. Write the amount given in the correct column
2Cu3N2 + 3O2---> 6CuO +*3.27 grams
2N2
4. Convert the amount given into moles
4. Convert the amount given into moles
2Cu3N2 + 3O2---> 6CuO +*3.27 grams*3.27g/1 x 1mole/79.545g= .0411 moles
2N2
5a. In each of the other columns write moles of
given (x) a fraction
5a. In each of the other columns write moles of
given (x) a fraction2Cu3N2 +*.0411mole x ___
3O2--->*.0411mole x ___
6CuO +*3.27 grams*3.27g/1 x 1mole/79.545g= .0411 moles
2N2*.0411 mole x ___
5b. The numerator of the fraction is the coefficient of
the column
5b. The numerator of the fraction is the coefficient of
the column2Cu3N2 +*.0411mole x 2/
3O2--->*.0411mole x 3/
6CuO +*3.27 grams*3.27g/1 x 1mole/79.545g= .0411 moles
2N2*.0411 mole x 2/
5c. The denominator of the fraction is the coefficient of
the given column
5c. The denominator of the fraction is the coefficient of
the given column2Cu3N2 +*.0411mole x 2/6
3O2--->*.0411mole x 3/6
6CuO +*3.27 grams*3.27g/1 x 1mole/79.545g= .0411 moles
2N2*.0411 mole x 2/6
5d. Do the math and label as moles
5d. Do the math and label as moles
2Cu3N2 +*.0411mole x 2/6=.0137 moles
3O2--->*.0411mole x 3/6=.0206 moles
6CuO +*3.27 grams*3.27g/1 x 1mole/79.545g= .0411 moles
2N2*.0411 mole x 2/6= .0137 moles
6. Convert all mole into grams
6. Convert all mole into grams
2Cu3N2 +*.0411mole x 2/6=.0137 moles*.0137mole/1 x 218.652g/1 mole=*3.00grams
3O2--->*.0411mole x 3/6=.0206 moles*.0206mole/1 x 31.998g/1 mole= *.659 grams
6CuO +*3.27 grams*3.27g/1 x 1mole/79.545g= .0411 moles
*3.27 grams
2N2*.0411 mole x 2/6= .0137 moles*.0137mole/1 x 28.0g/1 mole= *.384grams
7. Verify the law of conservation of mass
7. Verify the law of conservation of mass
2Cu3N2 +*.0411mole x 2/6=.0137 moles*.0137mole/1 x 218.652g/1 mole=*3.00grams
3.00g +
3O2--->*.0411mole x 3/6=.0206 moles*.0206mole/1 x 31.998g/1 mole= *.659 grams
.659g = 3.66grams
6CuO +*3.27 grams*3.27g/1 x 1mole/79.545g= .0411 moles
*3.27 grams
3.27g +
2N2*.0411 mole x 2/6= .0137 moles*.0137mole/1 x 28.0g/1 mole= *.384grams
.384g = 3.65grams
AnswerAnswer
• You would need 3.66 total grams of Copper II Nitride and Oxygen to produce 3.27 grams of Copper II Oxide.
• You would need 3.66 total grams of Copper II Nitride and Oxygen to produce 3.27 grams of Copper II Oxide.
