steps in doing a stoichiometry problem by: devin campbell

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.