Stoichiometry

Stoichiometry Stoichiometry – the process of using a balanced chemical equation to calculate the relative amounts of reactants and products in a reaction.

Recap Stoichiometry problems Remember: 1. Convert to moles (if necessary) 2. Use mole ratio from balanced chemical equation to find moles of other substance. 3. Convert moles back to mass, volume, or representative particles. (if necessary)

Example: mole-mole problem Calculate the number of moles of water produced by the reaction of 3.55 mol of O 2 with an excess of hydrogen. 2H 2 + O 2  2H 2 O 3.55 mol O 2 × 2 mol H 2 O 1 mol O 2 1 = 7.10 mol H 2 O

C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) What mass of oxygen, O 2, will be required to react exactly with 19.1g of propane, C 3 H 8, in the above reaction? 19.1 g C 3 H mol C 3 H g C 3 H 8 5 mol O 2 1 mol C 3 H mol O g O 2 ××× = 69.3 g O 2 Example: Mass – Mass Problem

2C 2 H 6 (g) + 7O 2 (g)  4CO 2 (g) + 6H 2 O(g) What mass of carbon dioxide, CO 2, will be produced in the complete combustion of 27.6g of ethane, C 2 H 6, in the above reaction? 27.6 g C 2 H mol C 2 H g C 2 H 6 4 mol CO 2 2 mol C 2 H mol CO g CO 2 ××× = 80.8 g CO 2

What volume of carbon dioxide, CO 2, will be produced from 290g of propane, C 3 H 8 and an excess of O 2 at STP. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g)

Percent Yield Actual yield Theoretical Yield × 100% = Percent Yield Note: usually you will first need to determine the theoretical yield using a stoichiometry calculation.

Example: What is the percent yield if 26.2 g of CaO is actually produced when the theoretical yield is 27.8g of CaO g 27.8 g × 100%= 94.2 %

Example: When 75.6g of SiO 2 is heated with an excess of C, the actual yield of SiC was 42.8 g. What is the percent yield of this reaction? SiO 2 + 3C  SiC + 2CO First determine theoretical yield by stoichiometry: 75.6 g SiO mol SiO g SiO 2 1 mol SiC 1 mol SiO g SiC 1.00 mol SiC1 ××× Theoretical Yield = 50.5 g 42.8g 50.5g × 100%= 84.8 %

Definitions The reactant that “runs out” or limits the reaction is called the limiting reactant. The reactant that still remains or there is extra is called the excess reactant.

Determining the limiting reactant Select one product and determine how many moles or mass each reactant will produce. Select one product and determine how many moles or mass each reactant will produce. The reactant that yields the lowest amount of this product is the limiting reactant. The reactant that yields the lowest amount of this product is the limiting reactant.Also: To find out how much excess reagent you must do a second stoichiometry calculation to figure out how much of the excess reagent was used. Then subtract the amount used from the initial amount.

Example: 2Cu + S  Cu 2 S If 50.0g of Cu reacts with 25.0g of S in the above reaction. What is the limiting reactant? 50.0 g Cu1.00 mol Cu g Cu 1 mol Cu 2 S 2 mol Cu1 ×× Cu =.393 mol Cu 2 S 25.0 g S1.00 mol S g S 1 mol Cu 2 S 1 mol S1 ××=.780 mol Cu 2 S S lowest amount of product => limiting reactant

Example: 2Cu + S  Cu 2 S If 50.0g of Cu reacts with 25.0g of S in the above reaction. What is the limiting reactant and the mass of Cu 2 S produced? 50.0 g Cu1.00 mol Cu g Cu 1 mol Cu 2 S 2 mol Cu1 ×× Cu 25.0 g S1.00 mol S g S 1 mol Cu 2 S 1 mol S1 × × = 62.6 g Cu 2 S S lowest amount of product is from limiting reactant g Cu 2 S × 1 mol Cu 2 S g Cu 2 S × 1 mol Cu 2 S = 124 g Cu 2 S

How many grams of rhodium (III) oxide are formed when 2.28 grams of rhodium are mixed with 0.56 grams of oxygen gas? What is the limiting reactant? 4Rh + 3O 2  2Rh 2 O g Rh mol Rh g Rh 2 mol Rh 2 O 3 4 mol Rh1.00 mol Rh 2 O g Rh 2 O 3 ××× = 2.81 g Rh 2 O 3.56 g O mol O g O 2 2 mol Rh 2 O 3 3 mol O mol Rh 2 O g Rh 2 O 3 ××× = 2.96 g Rh 2 O 3 Actual amount produced Rh is the limiting reactant

How many grams of water are formed when 3.46 grams of methane are mixed with 3.89 grams of oxygen gas? What is the limiting reactant? CH 4 + 2O 2  CO 2 + 2H 2 O 3.48 g CH mol CH g CH 4 2 mol H 2 O 1 mol CH mol H 2 O g H 2 O ××× = 7.77 g H 2 O 3.89 g O mol O g O 2 2 mol H 2 O 2 mol O mol H 2 O g H 2 O ××× = 2.19 g H 2 O Actual amount produced; O 2 is limiting