Units 8 & 9 Test Review – Chemistry I

Topics Covered on the Test:

* Balancing Equations                                                * Stoichiometry

* Types of Reactions                                       * Percent Yield

* Single Replacement Prediction                    * Limiting Reactant (Honors Only)

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1.)  ___ CaCl2  +  _2_ AgNO3  à  ___ Ca(NO3)2  +  _2_AgCl
50.0 g                                                                      ? g

(A) Balance the equation.

(B) Type of reaction? Double replacement

(C) If 50.0 grams of calcium chloride are reacted with excess silver nitrate, how many grams
of silver chloride can be produced?
50.0 g CaCl2 | 1 mole = 0.45 moles CaCl2
|111.1 g
0.45 moles CaCl2 = x moles AgCl                 x = 0.9 moles AgCl
1                            2
0.9 moles AgCl | 143.4 g = 129 g AgCl
| 1 mole

(D) What is the percent yield if a student makes 118 grams of silver chloride in this
experiment?

118 g x 100 = 91.5 % yield
129 g

2.)  _2_ Fe(HCO3)3  à  ___ Fe2O3  +  _3_ H2O  +  _6_CO2
15.0 g                       ? g

(A) Balance the equation.

(B) Type of reaction? Decomposition

(C) A lab group decomposed 15.0 grams of Fe(HCO3)3.  What is the theoretical yield of iron
(III) oxide?
15.0 g Fe(HCO3)3 | 1 mole = 0.0628 moles Fe(HCO3)3
|238.8 g
0.0628 moles Fe(HCO3)3 = x moles Fe2O3               x = 0.0314 moles Fe2O3
2                                      1
0.0314 moles Fe2O3 | 159.6 g = 5.01 g Fe2O3
| 1 mole

(D) If the lab group produced 4.63 grams of iron (III) oxide, what is their percent yield?
4.63 g x 100 = 92.4 % yield
5.01 g

3.)  ___ Au  +  ___HNO3  +  _3_HCl  à  ___ AuCl3  +  ___NO  +  _2_ H2O
1.25 moles                             ? g

(A) How many grams of hydrochloric acid (HCl) are needed to completely react 1.25 moles of
gold metal?
1.25 moles Au = x moles HCl                        x = 3.75 moles HCl
1                       3
3.75 moles HCl | 36.5 g = 137 g HCl
| 1 mole

4.)  ___ Fe  +  _2_ CuNO3  à  ___ Fe(NO3)2  +  _2_ Cu
3.7 moles                                                             ? moles

(A) How do we know that this reaction actually happens?
Iron (Fe) is higher on the Activity Series than copper (Cu).

(B) When 3.7 moles of iron are reacted with excess copper (I) nitrate, how many moles of
copper are produced?

3.7 moles Fe = x moles Cu                 x = 7.4 moles Cu
1                      2

5.)  ___ Ca3(PO4)2  +  _3_ SiO2  +  _5_ C  à  _2_ P  +  _3_ CaSiO3  +  _5_ CO
225.5 g                                                                          ? moles

(A) How many moles of CaSiO3 would be produced by the complete reaction of 225.5 grams of
calcium phosphate?
225.5 g Ca3(PO4)2 | 1 mole = 0.7267 moles Ca3(PO4)2
|310.3 g
0.7267 moles Ca3(PO4)2 = x moles CaSiO3             x = 2.810 moles Ca3(PO4)2
1                                      3

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Determine if the following reactions will occur.  (Honors:  If the reaction occurs, write the formula(s) for the product(s) and balance the equation.)

6.)       2 Al     +         6 HCl  à  2 AlCl3  +  3 H2

7.)       Zn(NO3)2        +         Mg       à  Zn  +  Mg(NO3)2

8.)       Cu       +         H2SO4             à  no reaction

9.)       CsF      +         I2         à  no reaction

10.)     Cl2       +         2 NaBr            à  Br2  +  2 NaCl

11.)     Pb(ClO3)2        +         2 K       à  Pb  +  2 KClO3

12.)     Mn      +         LiOH   à  no reaction

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Honors only

13.)  _2_ K3PO4  +  _3_ MgCl2  à  ___ Mg3(PO4)2  +  _6_ KCl
25.4 g               21.7 g                                               ? g

(A) What is the theoretical yield of potassium chloride if 21.7 grams of magnesium chloride
are reacted with 25.4 grams of potassium phosphate?
*First, find the limiting reactant (LR).*
25.4 g K3PO4 | 1 mole = 0.1196 moles K3PO4 = 0.0598*  - smaller # = LR
|212.3 g
2
25.4 g MgCl2 | 1 mole = 0.2277 moles MgCl2
= 0.0759

| 95.3 g                 3

0.1196 moles K3PO4 = x moles KCl              x = 0.3588 moles KCl
2                             6
0.3588 moles KCl | 74.6 g = 26.8 g KCl
| 1 mole

(B) How many grams of excess reactant remain after the reaction is complete?

0.1196 moles K3PO4 = x moles MgCl2                     x = 0.1794 moles MgCl2
2                             3
0.1794 moles MgCl2 | 95.3 g = 17.1 g MgCl2 used
| 1 mole

21.7 g MgCl2 at start

- 17.1 g MgCl2 used

4.6 g remaining