# Stoichiometry Ratios The stoichiometric coefficients in a balanced chemical reaction can be used to determine the mole relationships between any combination.

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Stoichiometry Ratios The stoichiometric coefficients in a balanced chemical reaction can be used to determine the mole relationships between any combination of the reactants and/or products. The types of ratios are: – Mole-Mole – Mass-Mole – Mass-Mass – Mass-Volume (at STP) – Volume - Volume (at STP)

Stoichiometry Practice Write the balanced equation for nitrogen reacts with hydrogen to produce ammonia gas. Write the balanced equation for nitrogen reacts with hydrogen to produce ammonia gas. A) If 1.47g of nitrogen reacts with an excess of hydrogen, how many grams of NH 3 are produced? A) If 1.47g of nitrogen reacts with an excess of hydrogen, how many grams of NH 3 are produced? B) How many grams of nitrogen would be required to produce 14.7 g NH 3 ? C) If 1.34 L of hydrogen reacted with an excess of nitrogen, how many liters of ammonia would be produced? If 18.5 g of silver nitrate reacts with an excess of sulfuric acid, how many grams of solid silver sulfate would be produced? If 18.5 g of silver nitrate reacts with an excess of sulfuric acid, how many grams of solid silver sulfate would be produced?

Percent Yield No chemical reaction is 100% efficient. No chemical reaction is 100% efficient. the amount of product produced is effected by impurities, side reactions, etc. the amount of product produced is effected by impurities, side reactions, etc. Theoretical yield is the maximum amount of product that can be produced. Theoretical yield is the maximum amount of product that can be produced. determined from stoich. problem determined from stoich. problem Actual yield is the amount of product produced. Actual yield is the amount of product produced. Percent Yield = Actual Yield Theoretical Yield X 100

Percent Yield Practice From the last example that you worked, if only 11.8 g Ag 2 SO 4 are actually produced, what is the percent yield? From the last example that you worked, if only 11.8 g Ag 2 SO 4 are actually produced, what is the percent yield?

Limiting Reagent In almost all chemical reactions, one of the reactants will run out first and limit how much product can be produced. The reactant that runs out first is the Limiting Reagent. In almost all chemical reactions, one of the reactants will run out first and limit how much product can be produced. The reactant that runs out first is the Limiting Reagent. Imagine you were making peanut butter and jelly sandwiches. You have 5 gallons of peanut butter, 5 gallons of jelly and 4 pieces of bread. How many sandwiches can you make? Imagine you were making peanut butter and jelly sandwiches. You have 5 gallons of peanut butter, 5 gallons of jelly and 4 pieces of bread. How many sandwiches can you make? Only 2. After making 2 sandwiches you still have plenty of the other ingredients, but you run out of bread. So the bread is the Limiting Reagent. Only 2. After making 2 sandwiches you still have plenty of the other ingredients, but you run out of bread. So the bread is the Limiting Reagent. When you don’t know which reactant is in excess, you use each reactant to calculate the moles of product it would produce. Whichever produces the least product is the limiting reagent and you always use this lesser amount of product to calculate the theoretical yield. When you don’t know which reactant is in excess, you use each reactant to calculate the moles of product it would produce. Whichever produces the least product is the limiting reagent and you always use this lesser amount of product to calculate the theoretical yield.

Limiting Reagent Think back to the example of hydrogen and nitrogen producing ammonia. Think back to the example of hydrogen and nitrogen producing ammonia. 3H 2 (g) + 1N 2 (g)  2NH 3 (g) If 12.1 g of nitrogen reacts with 10.4 g of hydrogen, how many grams of ammonia should be produced? If 12.1 g of nitrogen reacts with 10.4 g of hydrogen, how many grams of ammonia should be produced? 12.1 g N 2 1 mol N 2 28.02 g N 2 x 2 mol NH 3 1 mol N 2 =0.864 mol NH 3 x 10.4 g H 2 x 1 mol H 2 2.02 g H 2 x 2 mol NH 3 3 mol H 2 =3.43 mol NH 3 0.864 mol NH 3 x 17.04 g NH 3 1 mol NH 3 =14.7 g NH 3

Amount of Excess Left Over From the previous problem, how many g of the H 2 (g) is left over when the reaction is complete? From the previous problem, how many g of the H 2 (g) is left over when the reaction is complete? Subtract the theoretical yield (in moles) of the LR from the theoretical yield of the ER Subtract the theoretical yield (in moles) of the LR from the theoretical yield of the ER 3.43 mol - 0.864 mol = 2.57 mol Now work backward from there to get the mass of H 2 : Now work backward from there to get the mass of H 2 : 2.57 mol NH 3 x 3 mol H 2 x 2.02g H 2 = 7.79g H 2 2 mol NH 3 1 mol H 2 2 mol NH 3 1 mol H 2 If the problem had asked for excess L what would you do differently? 3H 2 (g) + 1N 2 (g)  2NH 3 (g)

The Grand Daddy of Them All!!! What is the percent yield if 12.7 g of copper (II) nitrate react with 6.11 g of sodium hydroxide and 4.70 g of solid copper (II) hydroxide are produced 12.7 g Cu(NO 3 ) 2 x 1 mol Cu(NO 3 ) 2 187.57 g Cu(NO 3 ) 2 x 1 mol Cu(OH) 2 1 mol Cu(NO 3 ) 2 = 0.0677 mol Cu(OH) 2 6.11 g NaOHx 1 mol NaOH 40.00 g NaOH x 1 mol Cu(OH) 2 2 mol NaOH = 0.0764 mol Cu(OH) 2 LR = Cu(NO 3 ) 2 and it yields.0677 mol Cu (OH) 2 Cu(NO 3 ) 2 (aq) + 2 NaOH(aq)  Cu(OH) 2 (s) + 2 NaNO 3 (aq)

Continued! 0.0677 mol Cu(OH) 2 x 97.57 g Cu(OH) 2 1 mol Cu(OH) 2 =6.61 g Cu(OH) 2 Percent Yield =x 100 4.70 6.61 =71.1% How much of the excess reagent is left over (in grams)? 0.0764 mol -0.0677 mol 0.0087 mol Cu(OH) 2 x 2 mol NaOH 1 mol Cu(OH) 2 x 40.00 g NaOH 1 mol NaOH = 0.70 g NaOH left over

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