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How to solve a stoichiometry problem By Erica Dougherty.

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Presentation on theme: "How to solve a stoichiometry problem By Erica Dougherty."— Presentation transcript:

1 How to solve a stoichiometry problem By Erica Dougherty

2 How many grams of calcium nitride and phosphorous are necessary to make 7.16g of calcium phosphide?

3 Write a complete and balanced chemical equation. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N

4 Draw a column after each chemical. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N

5 Write the amount given in the correct column. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N 7.16g

6 Convert the amount given into moles. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N 7.16g* 1mole / 3(40.08)+2(30.97)g =.0393 moles

7 Find moles for each of the other chemicals. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N 7.16g* 1mole / 3(40.08)+2(30.97)g =.0393 moles.0393 moles

8 In each of the other columns, write moles of a given times a fraction. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N 7.16g* 1mole / 3(40.08)+2(30.97)g =.0393 moles.0393 moles * /.0393 moles * /.0393 moles * /

9 The numerator is the coefficient of that column. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N 7.16g* 1mole / 3(40.08)+2(30.97)g =.0393 moles.0393 moles * 1/.0393 moles * 2/.0393 moles * 2/

10 The denominator is the coefficient of the given column. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N 7.16g* 1mole / 3(40.08)+2(30.97)g =.0393 moles.0393 moles * (1/1) =.0393 moles * (2/1) =.0393 moles * (2/1) =

11 Do the math and label as moles. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N 7.16g* 1mole / 3(40.08)+2(30.97)g =.0393 moles.0393 moles * (1/1) =.0393 moles.0393 moles * (2/1) =.0786 moles.0393 moles * (2/1) =.0786 moles

12 Convert all moles into grams. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N 7.16g* 1mole / 3(40.08)+2(30.97)g =.0393 moles 7.16 g.0393 moles * (1/1) =.0393 moles * (3(40.08) + 2(14.0067)) / 1mole = 5.83 g.0393 moles * (2/1) =.0786 moles * (30.974g / 1 mole) = 2.43 g.0393 moles * (2/1) =.0786 moles * 14.0067g/1mole = 1.10 g

13 Verify law of conservation of mass Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N 7.16g* 1mole / 3(40.08)+2(30.97)g =.0393 moles 7.16 g.0393 moles * (1/1) =.0393 moles * (3(40.08) + 2(14.0067)) / 1mole = 5.83 g.0393 moles * (2/1) =.0786 moles * (30.974g / 1 mole) = 2.43 g.0393 moles * (2/1) =.0786 moles * 14.0067g/1mole = 1.10 g 5.83g + 2.43g = 8.26g 7.16g + 1.10g = 8.26g

14 Answer 5.83 g of calcium nitride and 2.43 g of phosphorous are necessary to make 7.16g of calcium phosphide.


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