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Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

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Presentation on theme: "Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300."— Presentation transcript:

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2 Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300 400 500 100 200 300 400 500 100 200 300 400 500

3 Mole Conversion for 100 points ¿How many moles are in 28 grams of CO 2 ? 28.0gCO 2 x 1 mol CO 2 = 0.636 mol CO 2 44.01gCO 2

4 Mole Conversion for 200 points How many moles of magnesium is 3.01 x 1022 atoms of magnesium? 3.01e22 atom Mg x 1 mole Mg = 0.0499 mol Mg 6.022e23 atoms Mg

5 Mole Conversion for 300 points Determine the volume, in liters, occupied by 0.030 moles of a gas at STP. 0.030 mol x 22.4 L = 0.67 L 1 mole

6 Mole Conversion for 400 points How many oxygen molecules are in 3.36 L of oxygen gas at STP? 3.36 L O 2 x 1 mole O 2 x 6.022e23 molecules = 9.03e22 22.4 L O 2 1 mole O 2

7 Mole Conversion for 500 point Find the mass in grams of 2.00 x 1023 molecules of F2. 2.00e23 mc F 2 x 1 mole F 2 x 38g F 2 = 12.6g F 2 6.022e23 mc F 2 1 mole

8 Percent Composition for 100 points Find the percent composition of CuBr 2 Cu = 63.55 63.55(1) x 100 = 28.45% 223.35 Br x2 =+ 159.8 79.9(2) x 100 = 71.55% 223.35 223.35

9 Percent Composition for 200 points Find the percent composition of NaOH Na = 22.99 (1)22.99 x 100 = 57.48% 40.00 O = 16.00(1)16.00 x 100 = 40.00% 40.00 H = + 1.01 (1)1.01 x 100 = 2.53% 40.00 40.00

10 Percent Composition for 300 points Find the percent composition of N 2 S 2 N x2 = 28.02 14.01(2) x 100 = 30.44% 92.04 S x2 =+64.02 32.01(2) x 100 = 69.56% 92.04 92.04

11 Percent Composition for 400 points Find the percent composition of KMnO 4 K = 39.10 (1)39.10 x 100 = 24.74% 158.04 Mn = 54.94(1)54.94 x 100 = 34.76% 158.04 O x4= + 64.00 (4)16.00 x 100 = 40.50% 158.04 158.04

12 Percent Composition for 500 points Find the percent composition of Al 2 (SO 4 ) 3 Al x2= 53.96 (2)26.98 x 100 = 15.78% 341.99 S x3 = 96.03(3)32.01 x 100 = 28.08% 341.99 O x12= + 192(12)16.00 x 100 = 56.14% 341.99 341.99

13 Empirical and Molecular Formulas for 100 points Find the empirical formula for a compound containing.783% Carbon,.196% Hydrogen, and.521% Oxygen..783g C x 1 mol C = 0.0652 = 2 12.01g C 0.0326.196g H x 1 mol H =.194= 6=C 2 H 6 O 1.01g H 0.0326.521g O x 1 mol O = 0.0326= 1 16.00g O 0.0326

14 Empirical and Molecular Formulas for 200 points The empirical formula for a substance is CH 2 O. If its molar mass is 180, what is the molecular formula C = 12.01 H x2 = 2.02180 = 6 = 6(CH 2 O) = C 6 H 12 O 6 O = +16.00 30.03 30.03

15 Empirical and Molecular Formulas for 300 points The empirical formula for a substance is C 3 H 7 NO 3. If its molar mass is 105.11, what is the molecular formula C x3 = 36.03 H x7 = 7.07 N x1 = 14.01 105.11 = 1 = 1(C 3 H 7 NO 3 ) = C 3 H 7 NO 3 O x3 = +48.00 105.11 105.11

16 Empirical and Molecular Formulas for 400 points Find the empirical formula for a compound containing 1.388g Carbon,.345g Hydrogen, and 1.850g Oxygen. 1.388g C x 1 mol C = 0.1156 = 1 12.01g C 0.1156.345g H x 1 mol H = 0.3416= 3=CH 3 O 1.01g H 0.1156 1.850g O x 1 mol O = 0.1156 = 1 16.00g O 0.1156

17 Empirical and Molecular Formulas for 500 points Find the empirical and molecular formula for a compound containing 11.39g Phosphorus and 39.12g Chloride. Its molar mass is 274.64g. 11.36g P x 1 mol P = 0.3677 = 1 30.97g P 0.3677 39.12g Cl x 1 mol Cl = 1.1035 = 3PCl 3 35.45g Cl 0.3677 P x1 = 30.97274.64 = 2 = 2(PCl 3 ) = P 2 Cl 6 Cl x3 = +106.35 137.32 137.32

18 Balancing Equation for 100 points Balance: _N 2 + _H 2  _NH 3 Balanced: 1N 2 + 3H 2  2NH 3

19 Balancing Equation for 200 points Balance: _KClO 3  _KCL + _O 2 Balanced: 2KClO 3  2KCL + 3O 2

20 Balancing Equation for 300 points Balance: _Na + _H 2 O  _NaOH + _H 2 Balanced: 2Na + 2H 2 O  2NaOH + 1H 2

21 Balancing Equation for 400 points Balance: _FeCl 3 + _NaOH  _Fe(OH) 3 + _NaCl Balanced: 1FeCl 3 + 3NaOH  1Fe(OH) 3 + 3NaCl

22 Balancing Equation for 500 points Balance: _C 8 H 18 + _O 2  _CO 2 + _H 2 O Balanced: 2C 8 H 18 + 25O 2  16CO 2 + 18H 2 O

23 Stoichiometry for 100 points Given this equation:2 KClO 3 ---> 2 KCl + 3 O 2, How many moles of O 2 can be produced by letting 12.00 moles of KClO 3 react? 12 mol KClO 3 x 3 mol O 2 = 18 mol O 2 2 mol KClO 3

24 Stoichiometry for 200 points Given this equation: 2K + Cl 2 ---> 2KCl, how many moles of KCl would be produced from 2.50g of K and an excess of Cl 2 2.50g K x 1 mol K x 2 mol KCl = 0.0639 mol KCL 39.10g K 2 mol K

25 Stoichiometry for 300 points Given this equation: 2NaClO 3  2NaCl + 3O 2, how many grams of O 2 would be produced from 12.0 moles of NaClO 3 12 mol NaClO 3 x 3 mol O 2 x 32.0g O 2 = 576g O 2 2 mol NaClO 3 1 mol O 2

26 Stoichiometry for 400 points Given this equation: Na 2 O + H 2 O ---> 2NaOH, How many grams of NaOH can be produced from 54.8 grams of Na 2 O and an excess of H 2 O reacting 54.8g Na 2 O x 1 mol Na 2 O x 2 mol NaOH x 40 g NaOH = 35.4 61.98g Na 2 O 2 mol KClO 3 1 mol NaOH

27 Stoichiometry for 500 points Given this equation: 8Fe + S 8 ---> 8FeS, How many grams of FeS can be produced from 24.5 grams of S and an excess of Fe reacting 24.5g S 8 x 1 mol S 8 x 8 mol FeS x 87.86 g FeS = 67.2 256.08g S 1 mol S 8 1 mol FeS

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