pp. 326 – 334

 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.

 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?

 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.

 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 = ?

 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.

 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!

 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.

 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]

 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).

 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 g/mol 2.02 g/mol R: m CH3OH = ?

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

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

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

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