Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions

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

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 = mol Mg 6.022e23 atoms Mg

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

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 = 9.03e L O 2 1 mole O 2

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

Percent Composition for 100 points Find the percent composition of CuBr 2 Cu = (1) x 100 = 28.45% Br x2 = (2) x 100 = 71.55%

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

Percent Composition for 300 points Find the percent composition of N 2 S 2 N x2 = (2) x 100 = 30.44% S x2 = (2) x 100 = 69.56%

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

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

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 = = g C g H x 1 mol H =.194= 6=C 2 H 6 O 1.01g H g O x 1 mol O = = g O

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 = H x2 = = 6 = 6(CH 2 O) = C 6 H 12 O 6 O =

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 , what is the molecular formula C x3 = H x7 = 7.07 N x1 = = 1 = 1(C 3 H 7 NO 3 ) = C 3 H 7 NO 3 O x3 =

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

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 g g P x 1 mol P = = g P g Cl x 1 mol Cl = = 3PCl g Cl P x1 = = 2 = 2(PCl 3 ) = P 2 Cl 6 Cl x3 =

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

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

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

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

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

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 moles of KClO 3 react? 12 mol KClO 3 x 3 mol O 2 = 18 mol O 2 2 mol KClO 3

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 g K x 1 mol K x 2 mol KCl = mol KCL 39.10g K 2 mol K

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

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 = g Na 2 O 2 mol KClO 3 1 mol NaOH

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 g FeS = g S 1 mol S 8 1 mol FeS