1. pp. 326 – 334

2.  If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make?  The limiting reagent is the reactant you run out of first.  The limiting reagent determines how much product you can make  The excess reagent is the one you have left over.

3.  Nitrogen and hydrogen molecules react to form ammonia: N 2 + 3H 2  2NH 3  Use Fig. 7 p. 330 to answer the following: a) Identify the limiting and excess reagents in the flask. b) What is the maximum number of ammonia molecules that can be made? c) How many molecules of excess reagent remain?

4.  Step 1: Write a balanced chemical equation listing given value(s) and required value(s)  Step 2: Determine the limiting reagent by using the amount (n) of one reactant to find the stoichiometric amount (n) of the other.  Step 3: Use the amount of limiting reagent to find the amount of required substance.

5.  Determine the amount of titanium metal produced when 2.8 mol of titanium(IV) chloride reacts with 5.4 mol of magnesium.  Step 1: TiCl 4 + 2Mg  Ti + 2MgCl 2 G: 2.8 mol 5.4 mol R: n Ti = ?

6.  Step 2: n Mg = 2.8 mol TiCl4 × (2 mol Mg ÷ 1 mol TiCl4 ) n Mg = 5.6 mol  Since 5.6 mol is more than the 5.4 mol given in the question, magnesium must be limiting  It doesn’t matter which of the two reactants you choose to do this for, it will give you the same limiting reagent.

7.  For example: n TiCl4 = 5.4 mol Mg × (1 mol TiCl4 ÷ 2 mol Mg ) n TiCl4 = 2.7 mol  Since 2.7 mol is less than the 2.8 mol given in the question, magnesium must be limiting  If the calculated amout is less than the given amout, it is execess; if it is more than the given amout, it is limiting!

8.  Step 3: n Ti = 5.4 mol Mg × (1 mol Ti ÷ 2 mol Mg ) n Ti = 2.7 mol P: When 2.8 mol of titanium(IV)chloride is combined with 5.4 mol of magnesium, 2.7 mol of titanium will be produced.

9.  A nitric acid spill is neutralized by adding sodium hydrogen carbonate, NaHCO3(s): HNO 3(aq) + NaHCO 3(s) → H 2 O (l) + CO 2(g) + NaNO 3(aq)  What amount of water is produced when 2.3 mol of nitric acid is combined with 2.0 mol of sodium hydrogen carbonate? [ans: 2.0 mol]

10.  This follows the same strategy as for any other stoichiometry problem involving masses.  The only difference is that you first determine which reactant is the limiting reagent, then use the mass of the limiting reagent to determine the masses of product(s).

12.  What mass of methanol can be produced from 9.80 g of carbon monoxide & 1.30 g of hydrogen?  Step 1: Balanced Chemical Equation CO (g) + 2H 2(g)  CH 3 OH (l) G: 9.80 g 1.30 g 28.01 g/mol 2.02 g/mol R: m CH3OH = ?

13.  Step 2: Convert mass of given substances to amount (n) of given substances n CO = 9.80 g ÷ 28.01 g/mol = 0.34988 mol n H2 = 1.30 g ÷ 2.02 g/mol = 0.64356 mol  Step 3: Determine the limiting reagent n CO = 0.64356 mol H2 × (1 mol CO ÷ 2 mol H2 ) n CO = 0.32178 mol Since the amount of CO present initially is greater than this amount, CO is the excess reagent.

14.  Step 4: Use the amount of the limiting reagent to find the amount of required substance. n CH3OH = 0.64356 mol H2 × (1 mol CH3OH ÷ 2 mol H2 ) n CH3OH = 0.32178 mol  Step 5: Convert amount (n) of required substance to mass of required substance. m CH3OH = 0.32178 mol × 32.05 g/mol m CH3OH = 10.3 g

15. P: When 9.80 g of carbon monoxide reacts with 1.30 g of hydrogen, 10.3 g of methanol will be produced.

16.  Read pp. 326 – 324  Answer the following questions: p. 330 # 2, 3 p. 335 # 1, 3a, 4, 6 – 10  Read investigation 7.5.1 “What Stopped the Silver?” pp. 342 – 343 Create a data table and be prepared to begin this experiment tomorrow!