Slide 1

How to solve a stoichiometry problem By Erica Dougherty Slide 2 How many grams of calcium nitride and phosphorous are necessary to make 7.16g of calcium phosphide? Slide 3 Write a complete and balanced chemical equation. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N Slide 4 Draw a column after each chemical. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N Slide 5 Write the amount given in the correct column. Ca 3 N 2 + 2P ---> Ca 3 P 2 + 2N 7.16g Slide 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 Slide 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 Slide 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 * / Slide 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/ Slide 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) = Slide 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 Slide 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 Slide 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 Slide 14 Answer 5.83 g of calcium nitride and 2.43 g of phosphorous are necessary to make 7.16g of calcium phosphide.