# Mass relationships in chemical reactions: Stoichiometry

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Mass relationships in chemical reactions: Stoichiometry
Chapter 10 Mass relationships in chemical reactions: Stoichiometry

To Review How many grams in a mole of CaSO4 Al2(SO4)3
Ca 1 x 40.1 g/mol = g S 1 x = 32.1 O 4 x = 64.0 136.2 g/mol Al2(SO4)3 Al 2 x 27.0 g/mol = S 3 x = O 12 x = 312.3 g/mol

2Al(s) + 3Br2(l) 1Al2Br6(s) Stoichiometric Coefficients- the coefficients (number in front of the chemical formula) in a balanced chemical equation Stoichiometry- the relationship between the quantities of chemical reactants and products Stoichiometric factor- a mole ratio relating moles of one substance to another substance involved in the same chemical reaction

2 H2O2(liq) ---> 2 H2O(l) + 1 O2(g)
Create two stoichiometric factors that relate the amount of the reactant H2O2 to the product O2 2 H2O2(liq) ---> 2 H2O(l) O2(g) 2 mol H2O OR mol O2 1 mol O molH2O2

2 H2O2(l) ---> 2 H2O(l) + O2(g)
Create two stoichiometric factors that relate the amount of the product H2O to the reactant H2O2 2 H2O2(l) ---> 2 H2O(l) + O2(g) 2 mol H2O2 OR mol H2O 2 mol H2O mol H2O2

8 mol CO2 OR 2 mol C4H10 2 C4H10 + 13O2 -----> 10H2O + 8CO2
Create two stoichiometric factors that relate the amount of the product CO2 to the reactant C4H10 2 C4H O > 10H2O CO2 8 mol CO OR mol C4H10 2 mol C4H mol CO2

Stoichiometric Calculations
The most common types of equation stoichiometry problems are called 3 step problems Given the mass of a reactant or product determine the equivalent mass of a different reactant or product These are called mass to mass or gram to gram conversions Consider: 2 H2O > 2 H O2 How many g of H2 are produced by decomposing 1.00 g H2O?

Stoichiometric Calculations
All calculations are done using conversion factors Mass H2O Mol H2O Mol H2 Mass H2 1 mol MM in g mol H2 mol H2O MM in g 1 mol Each arrow represents a conversion factor Align units first, then place numbers into the conversion factor

Stoichiometric Calculations
Step 1 - Convert g of given reactant or product to moles Use molar mass of the given Step 2 - Convert from moles of the given to moles of the unknown Use stoich factor as conversion factor Step 3 - Convert from moles to g of the unknown Use molar mass of the unknown as conversion factor

Problem 1a. If 37.6 g of water is decomposed to hydrogen and oxygen, how many grams of hydrogen will be produced? Write the balanced chemical equation 2 H2O > 2 H O2

Write the information that is given and what's asked for above the equation
Mass 37.6 g ? g Equation 2H2O > 2H O2

Write the molar mass of the substances.
Mass 37.6 g ? g Equation 2 H2O > 2H O2 Molar Masses (g/mol) Remember the Pathway: g H2O mol H2O mol H2 g H2

Step 1 - Convert the given mass of water to moles of water
2 H2O > 2 H O2 1 mol H2O = 18.0 g H2O 37.6 g H2O x 1 mol H2O 18.0 g H2O

Step 2 - Convert moles of water to moles of hydrogen using stoichiometric factor
2 H2O > 2 H O2 37.6 g H2O x 1 mol H2O x 2 mol H2 18.0 g H2O 2 mol H2O

Step 3 - Convert moles of hydrogen to mass of hydrogen
2 H2O > 2 H O2 1 mol H2 = g H2 37.6g H2O x 1 mol H2O x 2 mol H2 x 2.0g H2 = 4.17g H2 18.0g H2O mol H2O 1 mol H2

Problem 1b. At room temperature and pressure, aluminum reacts with oxygen to give aluminum oxide. If you react 161 g of Al, what mass of O2 is needed for complete reaction? Write the chemical equation: Al(s) + O2(g) > Al2O3

You need a balanced chemical equation:
Al(s) O2(g) > Al2O3 2 4 3 2

Record given mass and what is asked for in the problem
Mass 161 g ? g Equation 4Al(s) + 3O2(g) ---->2Al2O3 Molar Masses (g/mole)

3 Step Calculations 4Al(s) + 3O2(g) ---->2Al2O3
g Al  mol Al  mol O2  g O2 161 g Al ? g O2 4Al(s) + 3O2(g) ---->2Al2O3 27.0 g/mol g/mol 161 g Al x 1 mol Al 27.0 g Al x 3 mol O2 4 mol Al x g O2 1 mol O2 = 143 g O2

What mass of MgO can be produced from igniting 1.5 g of Mg in oxygen?
Mass g ? g Equation 2 Mg(s) + O2(g) ---> 2 MgO Molar Masses (g/mole) 1.5 g Mg x 1 mol Mg x 24.3 g Mg 2 mol MgO x 2 mol Mg 40.3 g MgO 1 mol MgO = 2.49 g MgO

Theoretical vs Actual Yield
Theoretical yield = the amount of product that could be produced if the reaction is perfect The amount that results from the 3 step process This never happens You usually get less than the theoretical yield Actual yield = The amount of product you recover when you run the experiment

Limiting and Excess Reactant
In chem reactions, reactants are never present in the exact ratio to ensure they are completely used up in the reaction You usually run out of one reactant, and the other is in excess (left over) Limiting reactant = the one completely used up Excess reactant = the one with some left over

A Chemical Reaction Reactants Products
All of is used up = Limiting Reactant Some of is left over = Excess Reactant

Limiting Reactant Problem
Determine the limiting and excess reactant if 1.40 g of N2 reacts with 1.00 g of H2 to produce ammonia, NH3. 1.40 g g N H2  2NH3 28.O g/mol g/mol 1.40 g N2 x 1 mol N2 28.0 g N2 x 3 mol H2 1 mol N2 x 2.0 g H2 1 mol H2 = 0.30 g H2 There is more than 0.30 g of H2 present, therefore H2 will be left over (is in excess) and N2 will be the limiting reactant

Percent Yield A means of quantifying how efficient a chemical reaction is Actual Yield Theoretical Yield % Yield = x 100 % Yields are usually below 100% less product is recover than is possible % Yields can be above 100% What’s better, a % yield of 90% or 110% ?

Solving Percent Yield Probs
1 – Determine the limiting reactant 2 – Calculate the theoretical yield from the given amount of the limiting reactant 3 – Calculate the % yield (the actual yield needs to be given to you in the problem). Actual Yield Theoretical Yield % Yield = x 100

Percent Yield Practice Prob #1
A student reacts 6.00 g of N2 reacts with g of H2 to produce ammonia, NH3. if 2.59 g of NH3 is recovered from the reaction, what is the student’s % yield? N H2  2NH3 Step 1 Determine the limiting reactant. 6.00 g N2 x 1 mol N2 28.0 g N2 x 3 mol H2 2 mol N2 x 2.0 g H2 1 mol H2 = g H2 Therefore, g H2 is the limiting reactant

Percent Yield Practice Prob #1
Step 2 Deter the theoretical yield from limiting reactant 0.500 g H2 x 1 mol H2 2.0 g H2 x 2 mol NH3 3 mol H2 x 17.0 g NH3 1 mol NH3 = 2.83 g NH3 Actual yield = 2.59 g NH3 (given in the problem) Theoretical yield = 2.83 g NH3 % yield = actual yield x 100 theroretical yield = g x = 91.5% 2.83 g

Lab Exercise A student reacts 2.40 g of NaHCO3 with an excess of HCL
according to the reaction: 2.40 g X g NaHCO3 + HCl ---> NaCl H2O + CO2 84.0 g/mol g/mol Determine the theoretical yield of NaCl (3 step process) g NaHCO3  mol NaHCO3  mol NaCl  g NaCl 2.40 g NaHCO3 x 1 mol NaHCO3 84.0 g NaHCO3 x 1 mol NaCl 1 mol NaHCO3 x g NaCl = 1 mol NaCl = 1.67 g NaCl

Lab Exercise The reaction from your lab is conducted with a 2.40 g sample of NaHCO3. In the lab, the reaction produces 1.57 g of NaCl. What is the percent yield? x 1 mol NaHCO3 84.0 g/mol x 1 mol NaCl 1 mol NaHCO3 x g NaCl = 1 mol NaCl 1.67 g NaCl 2.40 g NaHCO3 % Yield = actual yield x 100 theoretical yield = g x 100 1.67 g = 94.0%