Chem 121 Ans to Extra Practice Problems Dealing with Moles, Grams, Stoichiometry

Fall 2006

This is not meant to be a tutorial, just some extra practice problems. If you are not getting the right

answers (including correct number of sig. fig.) you need to get help, by coming to see me or a tutor.

1. How many H atoms are in 78.9 g of C6H14?

ANS. 7.72x1024 H

(MM(C5H14) = 86.17 g/mol)

2. How many anions are in 375 mg of CaCl2?

ANS. 4.07x1021 anions

(MM(CaCl2) = 111.0 g/mol)

3. How many ions are in 375 mg of CaCl2? ANS. 6.10x1021 ions

4. How many moles of cations are in 375 mg of CaCl2? ANS. 3.38x10-3 mol cations

5. How many grams are in 583 molecules of CO2? ANS. 4.26x10-20 g CO2

6. Fermentation is a complex chemical process of wine making in which glucose is converted into

ethanol and carbon dioxide:

C6H12O6 ¾® 2C2H5OH + 2CO2

Starting with 500.4 g of glucose, what is the theoretical yield of ethanol?

ANS. 256 g ethanol

(MM=180.2 g/mol)

7. When CuSO4–5H2O (which is blue) is heated, it loses its water molecules and becomes white:

CuSO4–5H2O ¾® CuSO4 + 5H2O

If 9.60 g of CuSO4 are left after heating 15.01 g of the blue compound, calculate the number of

moles of water originally present in the compound.

Think carefully before you start this problem. There is a very easy solution to this problem.

ANS. 0.300 mol water

8. Consider the reaction

MnO2 + 4HCl ¾® MnCl2 + Cl2 + 2H2O

If 0.86 mole of MnO2 and 48.2 g of HCl react, which reagent will be used up first? How many g of

Cl2 will be produced? Before you look at the answers, reread the question. Have you answered

ALL the parts in this question?

ANS. HCl, 23.4 g Cl2

9. How many grams of each product would you expect from the following reaction, and how many

grams of which reactant is left over if we start with 2.65 g of barium chloride and 6.78 g of H2SO4.

If 1.25 g of BaSO4 is obtained, what is the percent yield?

BaCl2 + H2SO4 ¾®BaSO4 + 2HCl

Again, reread the question to make sure you have answered ALL the parts of this question.

ANS. 2.97 g BaSO4, 0.928 g HCl, 5.53 g H2SO4, 42.1 % yield

(MM(BaCl2) = 208.2 g/mol MM(H2SO4) = 98.09 g/mol MM(BaSO4) = 233.3 g/mol

MM(HCl) = 36.46 g/mol)

10. Cisplatin [Pt(NH3)2Cl2], a compound used in cancer treatment, is prepared by reaction of ammonia

with potassium tetrachloroplatinate:

K2PtCl4 + 2NH3 ¾®2 KCl + Pt(NH3)2Cl2

How many g of cisplatin are formed from 55.8 g of K2PtCl4 and 35.6 g of NH3 if the reaction takes

place in 95% yield.

ANS. 38.3 g cisplatin MM(K2PtCl4)=415.1g/mol MM(NH3)=17.03g/mol

MM(cisplatin)=300.1g/mol

If you can’t figure out why your answer is different from mine, click here.