PRACTICE FOR QUIZ – 9/18/06

Below are the answers to some problems in the book. TRY OUT THE PROBLEMS FIRST, and then check your answer. For number 71, ignore the question about how much of reactant is left over…

Limiting Reactants: 3.71, 3.75

3.71: How many grams of each product result from the reactions?

a. 1.3 g NaCl + 3.5 g AgNO3 → AgCl + NaNO3

Molar Mass of NaCl: 58.44 g/mol; AgNO3: 169.9 g/mol; AgCl: 143.3 g/mol; NaNO3: 85.00 g/mol

1.3 g NaCl x 1 mol NaCl = 0.022 mol NaCl

58.44 g NaCl

3.5 g AgNO3 x 1 mol AgNO3 = 0.021 mol AgNO3

169.9 g AgNO3

Since the equation tells us we only need 1 mol NaCl per mol AgNO3, we see that AgNO3 is limiting because it is less than NaCl and NaCl is in excess.

0.021 mol AgNO3 x 1 mol AgCl x 143.3 g AgCl = 3.0 g AgCl

1 mol AgNO3 1 mol AgCl

0.021 mol AgNO3 x 1 mol NaNO3 x 85.00 g NaNO3 = 1.8 g NaNO3

1 mol AgNO3 1 mol NaNO3

b. _1_ BaCl2 + _1_ H2SO4 → _1_BaSO4 + _2_HCl ; 2.65 g BaCl2, 6.78 g H2SO4

2.65 g BaCl2 x 1 mol BaCl2 = 0.0127 mol BaCl2

208.2 g BaCl2

6.78 g H2SO4 x 1 mol H2SO4 = 0.0691 mol H2SO4

98.08 g H2SO4

Comparing moles in a ratio: 0.0691 mol H2SO4 = 5.44 mol H2SO4 per 1 mol BaCl2

0.0127 mol BaCl2

Since the mole ratio is (again) 1 to 1, H2SO4 is in excess and BaCl2 is limiting. To find the grams of products obtained:

0.0127 mol BaCl2 x 1 mol BaSO4 x 233.4 g BaSO4 = 2.96 g BaSO4

1 mol BaCl2 1 mol BaSO4

0.0127 mol BaCl2 x 2 mol HCl x 36.46 g HCl = 0.926 g HCl

1 mol BaCl2 1 mol HCl

3.75: K2PtCl4 + 2NH3 → 2KCl + Pt(NH3)2Cl2

How many grams of cisplatin, Pt(NH3)2Cl2, are formed from 55.8 g of K2PtCl4 and 35.6 g NH3 if the reaction takes place in 95% yield based on the limiting reactant?

Molar Mass of K2PtCl4: 415.1 g/mol; NH3: 17.03 g/mol; Pt(NH3)2Cl2: 300.0 g/mol

55.8 g K2PtCl4 x 1 mol K2PtCl4 = 0.134 mol K2PtCl4

415.1 g K2PtCl4

35.6 g NH3 x 1 mol NH3 = 2.09 mol NH3

17.03 g NH3

Since you need 2 mol of NH3 per 1 mole of K2PtCl4, we know that we only need 2 x (0.134 mol K2PtCl4) = 0.268 mol K2PtCl4. Since we have less than 0.268 mol K2PtCl4, we know that K2PtCl4 is the limiting reagent and NH3 is therefore in excess. (Or you can look at it as a ratio. 2.09 mol NH3/0.134 mol K2PtCl4 = 15.6 mol NH3/1 mol K2PtCl4. From looking at the equation saying that we need only 2—and not 15.6 mol—per 1 mol of K2PtCl4, we see that NH3 is in excess and K2PtCl4 is the limiting reagent!)

0.134 mol K2PtCl4 x 1 mol Pt(NH3)2Cl2 x 300.0 g Pt(NH3)2Cl2 = 40.2 g Pt(NH3)2Cl2

1 mol K2PtCl4 1 mol Pt(NH3)2Cl2

Remember, this reaction is only 95% yield; therefore, 40.2 g Pt(NH3)2Cl2 is our theoretical yield. We need to find our actual yield—how much cisplatin we actually got in the lab (theoretical yield is always calculated, and actual is gotten in lab).

% Yield = Actual/Theoretical x 100 → 95% = Actual/40.2 g Pt(NH3)2Cl2 x 100 → Actual = 38 g Pt(NH3)2Cl2

% Composition: 3.95

What are the empirical formulas of each of the following substances? (First step: assume that each percent is in grams!)

c. Zircon: 34.91% O, 15.32% Si, 49.76% Zr

34.91 g O x 1 mol O = 2.182 mol O

16.00 g O

15.32 g Si x 1 mol Si = 0.5454 mol Si

28.09 g Si

49.76 g Zr x 1 mol Zr = 0.5455 mol Zr

91.22 g Zr

Divide each mol by the smallest amount—0.5454 mol Si.

Zr0.5455/0.5454Si0.5454/0.5454O2.182/0.5454 = ZrSiO4