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# Stoichiometry TJ Bautista, Sean Higgins, Joanna Lee Period 6.

## Presentation on theme: "Stoichiometry TJ Bautista, Sean Higgins, Joanna Lee Period 6."— Presentation transcript:

Stoichiometry TJ Bautista, Sean Higgins, Joanna Lee Period 6

Definitions Limiting Reactant/Limiting Reagent: the reactant that limits the amount of the other reactants that can combine and the amount of product that can form in the chemical reaction. Excess reactant: The substance that is not used up completely in a reaction

How to Solve for Limiting Reagent 1. Write the balanced chemical equation 2. Determine the moles of each reactant 3. Determine how many moles of product EACH reactant would make using a mole ratio 4. The reactant that yields less products is the limiting reagent

How does the limiting reactant affects the amount of products formed? The limiting reactant is the reactant that produces the smallest number of moles after all the calculations are performed

1) If 10.0 grams of NaOH react with 20.0 grams of H 2 SO 4 to produce Na 2 SO 4, which reactant is limiting? NaOH + H 2 SO 4 --> Na 2 SO 4 + H 2 O

1) If 10.0 grams of NaOH react with 20.0 grams of H 2 SO 4 to produce Na 2 SO 4, which reactant is limiting? NaOH + H 2 SO 4 --> Na 2 SO 4 + H 2 O 10.0g NaOH x 1 mole NaOH x 1 mole Na 2 SO 4 = 0.125 mole Na 2 SO 4 40g NaOH 2 mole NaOH 20.0g H 2 SO 4 x 1 mole H 2 SO 4 x 1 mole Na 2 SO 4 = 0.204 mole Na 2 SO 4 98.09g H 2 SO 4 1 mole H 2 SO 4 NaOH is the limiting reactant

2) If 15.0 grams of copper metal react with a solution containing 10.0 grams of AgNO 3, which reactant is limiting? Cu + AgNO --> Cu(NO ) + Ag 3 3 2

2) If 15.0 grams of copper metal react with a solution containing 10.0 grams of AgNO 3, which reactant is limiting? Cu + 2AgNO --> Cu(NO ) + 2Ag 3 3 2 15.0g Cu x 1 mole Cu x 2 mole Ag = 0.472 mole Ag 63.54g Cu 1 mole Cu 10.0g AgNO 3 x 1 mole AgNO 3 x 2 mole Ag = 0.0589 mole Ag 169.91g AgNO 3 2 mole AgNO 3 AgNO 3 is the limiting reactant

3) A 2.00 grams sample of ammonia is mixed with 4.00 of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped? NH 3 (g) + O 2 (g) --> NO(g) + H 2 O(g)

3) A 2.00 grams sample of ammonia is mixed with 4.00 of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped? 4NH 3 (g) + 5O 2 (g) --> 4NO(g) + 6H 2 O(g)

2.00g NH 3 x 1 mole NH 3 x 4 mole NO x 30.0g NO = 3.53 g NO 17.0g NH 3 4 mole NH 3 1 mole NO 4.00g O 2 x 1 mole O 2 x 4 mole NO x 30.0g NO = 3.00 g NO 32.0g O 2 5 mole O 2 1 mole NO **O 2 is the limiting reactant** 4.00g O 2 x 1 mole O 2 x 4 mole NH 3 x 17.0g NH 3 = 1.70 g NH 3 32.0g O 2 5 mole O 2 1 mole NH 3 2.00g NH 3 (original sample) - 1.70g (reacted) = 0.30g NH 3 (remaining) amount of ammonia that reacted

4) You have 22 grams of sodium nitrate and 16 g of sulfuric acid. Which reactant is in excess? How many grams is that reactant in excess? NaNO 3 (s) + H 2 SO 4 (l) --> Na 2 SO 4 (s) + HNO 3 (g)

4) You have 22 grams of sodium nitrate and 16 g of sulfuric acid. Which reactant is in excess? How many grams is that reactant in excess? 2NaNO 3 (s) + H 2 SO 4 (l) --> Na 2 SO 4 (s) + 2HNO 3 (g)

22g NaNO 3 x 1 mole NaNO 3 x 1 mole H 2 SO 4 x 98g H 2 SO 4 84.9g NaNO 3 2 mole NaNO 3 1 mole H 2 SO 4 = 12.6g H 2 SO 4 **NaNO 3 is the limiting reactant** 16g Sulfuric Acid - 12.6g Sulfuric Acid = 3.4g Sulfuric Acid (remaining)

Definitions Theoretical Yield: is the maximum amount of product that can be produced from a given amount of reactant Actual Yield: the measured amount of a product obtained from a reaction Percentage Yield: is the ratio of the actual yield to the theoretical yield times 100

Percentage Yield percentage = actual yield yield theoretical yield X 100

5) When 40.0 grams C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 50g. What is the percentage yield of C 6 H 5 Cl? C 6 H 6 (l) + Cl 2 (g) --> C 6 H 5 Cl(l) + HCl(g)

5) When 40.0 grams C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 50g. What is the percentage yield of C 6 H 5 Cl? C 6 H 6 (l) + Cl 2 (g) --> C 6 H 5 Cl(l) + HCl(g) 40.0g C 6 H 6 x 1 mole C 6 H 6 x 1 mole C 6 H 5 Cl x 112.56g C 6 H 5 Cl 78.12g C 6 H 6 1 mole C 6 H 6 1 mole C 6 H 5 Cl = 57.6g C 6 H 5 Cl theoretical yield percentage yield = 40.0g X 100 = 69.4% 57.6g

6) If 80.0 grams CO reacts to produce 70.0 grams CH 3 OH, what is the percentage yield of CH 3 OH? CO(g) + 2H 2 (g) -------------> CH 3 OH(l) catalyst

6) If 80.0 grams CO reacts to produce 70.0 grams CH 3 OH, what is the percentage yield of CH 3 OH? CO(g) + 2H 2 (g) -------------> CH 3 OH(l) catalyst 80.0g CO x 1 mole CO x 1 mole CH 3 OH x 32.04g CH 3 OH 28.01g CO 1 mole CO 1 mole CH 3 OH = 91.5g CH 3 OH percentage yield = 80.0g X 100 = 87.4% 91.5g theoretical yield

Definitions Empirical Formula: gives the simplest whole- number ratio of the atoms of the elements Molecular Formula: shows the types and numbers of atoms combined in a single molecular compound

7) Find the empirical formula of a compound containing: 19.32 % Ca, 34.30 % Cl, and 46.38 % O.

19.32g Ca x 1 mole Ca = 0.48204 mole Ca = 1 40.08g Ca 0.48204 34.30g Cl x 1 mole Cl = 0.96756 mole Cl = 2 35.45g Cl 0.48204 46.38g O x 1 mole O = 2.89875 mole O = 6 16.00g O 0.48204 mole ratio 1:2:6 Empirical formula = CaCl 2 O 6 --> Ca(ClO 3 ) 2

8) Find the empirical formula of a compound containing: 64.86 % C, 13.52 % H, and 21.62 % O.

64.86g C x 1 mole C = 5.4005 mole C = 1 12.01g Ca 1.35125 13.52g H x 1 mole H = 13.38614 mole Cl = 2 1.01g H 1.35125 21.62g O x 1 mole O = 1.35125 mole O = 6 16.00g O 1.35125 mole ratio 1:2:6 Empirical formula = C 4 H 10 O

9) Determine the molecular formula for a compound whose empirical formula is C 5 H 7 and whose molecular mass is 268.44 g/mole.

formula mass C 5 H 7 = 67.11 g/mole 268.44 g/mole = 4 67.11 g/mole 4 x (C 5 H 7 ) = C 20 H 28

10) Find the molecular formula for a compound that contains 42.56g of palladium and 0.80 of hydrogen. The molecular mass of the compound is 433.68 g/mole.

42.56g Pd x 1 mole Pd = 0.4 mole Pd = 1 106.4g Pd 0.4 PdH 2 0.80g H x 1 mole H = 0.79 mole H = 2 1.01g H 0.4 433.68 g/mole = 4 --------------> 4 x (PdH 2 ) = Pd 4 H 8 108.42 g/mole *Formula mass PdH 2 = 108.42 g/mole

That concludes this presentation of Chapter 9 Stoichiometry! (: http://oformi.net/uploads/gallery/main/31/12319.jpg

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