Answers to Exam 2 Spring 2000 Chem 111

1. A compound is composed of only C and H. A sample contains 2.38 g C and 0.200 g H. What is the empirical formula of this compound.

Solution: Divide the mass of C by 12 g/mol and the mass of H by 1 g/mol to get moles.

```
2.38                0.200
---- = 0.198        ----- = 0.200
12                   1

Note that 0.198  is equivalent to  1.0
-----                    ---
0.200                    1.0

```
Thus the formula has 1 C to 1 H. 2. A substance has an empirical formula of P2O3. The molar mass is 220 g/mol. What is it's molecular formula?

Solution: determine the mass of the empirical formula.

2(31) + 3(16) = 110

Divide the molar mass by the empirical mass to get multiplier.

```
220
--- = 2
110
```
Multiply the empirical formula by the multiplier:

2(P2O3) = P4O6

3. What is the %C by mass in acetone: C3H6O?

Solution: Determine the total mass.

12(3) + 1(6) + 16(1) = 58

The determine the C.

12(3) = 36

Divide the C mass by the total mass and multiply by 100.

```
36
-- x 100 = 62.1
58
```
Reaction: 2Na + Cl2 --> 2NaCl

4. What is the mass of Cl2 needed to react with 4.6g Na?

Solution: Determine moles of Na in 4.6g.

```
4.6
--- = 0.20
23
```
Note stoichiometry of 2 moles Na to 1 mole Cl2 so:

```
2   0.20
- = ----   x = 0.1
1    x
```
Multiply the mass of Cl2 by 0.1:

0.1 X 71 = 7.1

5. What mass of NaCl can be produced from 4.6 g Na and an excess of Cl2?

Solution: Use Law of Conservation of Mass

```
Given:                 4.6 g Na
From 4:                7.1 g Cl2
---
11.7 g NaCl
```
6. A reaction mixture contains 5.0 mol Na and 4.0 mol Cl2. How many moles of NaCl can be produced?

Solution: look at reaction stoichiometry.

2Na + Cl2 --> 2NaCl

2 Na -> 2 NaCl

```
Setup up ratio:     2     5
-  =  -          x = 5
2     x

```
7. How many moles of which reagent will be left unreacted from the mixture in the previous question?

Solution: Look at reaction stoichiometry.

2Na + Cl2 --> 2NaCl

2Na to 1Cl2

Determine how much Cl2 is used.

```
Setup up ratio:    2    5
- =  -       x = 2.5
1    x
```
2.5 moles used so 4.0 - 2.5 = 1.5 mol Cl2 left over.

8. What mass of NH3 will be produced from the reaction

N2(g) + 3H2(g) --> 2NH3(g)

if we started with 14.0g N2 and 12.0g H2?

Solution: look at stoichiometry.

1 N2 to 3 H2 -> 2 NH3

Determine moles N2 and moles H2.

```
14g                    12
---  = 0.5 mole        -- = 6 moles
28g                     2
```
Note have 6:0.5 which is 12 mol H2 to 1 mol N2 but reaction only needs 3:1. There is excess H2.

From stoichiometry:

```
1 N2    0.5 N2
----- = ------     x = 1 NH3
2 NH3    x NH3
```
Determine mass of 1 NH3: 14 + 3(1) = 17 g

9. For the reaction:

2Al + Fe2O3 --> 2Fe + Al2O3

the calculated yield was 185 g Fe. Only 162g Fe were recovered. What is the percent yield of thies reaction?

Solution: divide actual yield by theoretical (calculated) yield and multiply by 100.

```
162g
---- x 100 = 88%
185g
```
10. What is the molarity of glucose solution if 54 g glucose, C6H12O6 (180g/mol) is dissolved in enough water to make 0.500 L of solution?

Solution: molarity is:

```
moles
-----
L
```
First convert 54g to moles by dividing by the molar mass.

```
54g
-------- = 0.30
180g/mol
```
Divide moles by L to get molarity:

```

0.30 moles
----------- = 0.60 M
0.500 L

```
11. A solution is prepared by diluting 25 mL of 6.0 mol/L HCl to 250 mL with water. What is the concentration of the dilute solution?

Solution: Use m1v1 = m2v2

(6.0 mol/L)(25 mL) = (x mol/L)(250 mL)

Note since mL occurs on both sides there is no need to convert mL to L since the units will cancel.

x mol/L = 0.60 mol/L

12. It took 30.0 mL of 0.200 mol/L NaOH to completely react with 50.0 mL of an HCl solution.

HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)

What is the concentration of the HCl solution?

Solution:

Reaction stoichiometry is 1 mole to 1 mole.

So use

m1v1 = m2v2

(0.200 mol/M)(30.0 mL) = (x mol/L)(50.0 mL)

x = 0.120 mol/L

13. Mg(OH)2 is insoluble in water.

Mg2+(aq) + 2OH-(aq) --> Mg(OH)2(s)

100 mL of 0.50 mol/L NaOH is mixed with 200 mL of 0.10 mol/L MgCl2. Which statement is true?

Solution: determine stoichiometry.

1 mole Mg2+ to 2 moles OH-

Determine moles of each reactant. NaOH releases 1 mole OH- from 1 mole NaOH.

```
100mL
--------- x 0.50 mol/L NaOH = 0.05 mol OH-
1000 mL/L

200 mL
--------- x 0.20 mol/L MgCl2 = 0.02 mol Mg2+
1000 mL/L

Ratio is:

1 Mg2+
------
2 OH-

We have:
0.02 Mg2+             1
---------  which is  ---
0.05 OH-             2.5
```
There is more OH- than needed to use up all the Mg2+ so Mg2+is limiting. Since Mg2+ comes from MgCl2, MgCl2 is the limiting reagent.

14. Which of the following is considered a weak acid?

Solutions: organic (based on C) are always weak acids. Since only one answer can be correct, it must be HCH3COO.

15. Which of the following solutions will form a precipitate when mixed?

Solution: Use solubility rules.

A) all chlorides (except noted) and nitrates are soluble. No precipitate.

B) All carbonates (except noted) are insoluble. Insoluble means doesn't dissolve. Doesn't dissolve means can't be in solution. Can't be in solution means it precipitates.

C) All sulfates (except noted) and all chlorides are soluble.

D) This is a noted carbonate and not a noted sulfate.

E) All chlorides (except noted) are soluble. All hydroxides except noted are insoluble. This is a noted hydroxide and is soluble.

16. Which of the following sets of coefficients properly balance the following precipitation reaction?

NaH2PO4(aq) + CaCl2(aq) --> Ca3(PO4)2(s) + NaCl(aq) + HCl(aq)

Solution: Balance reaction.

2NaH2PO4(aq) + 3CaCl2(aq) --> Ca3(PO4)2(s) + 2NaCl(aq) + 4HCl(aq)

17. A weak acid is placed in a beaker and a pair of electrodes, connected to an electrical outlet on one end and a light bulb on the other, are placed in the beaker. Power is appleid to the "circuit". What happens?

Solution:

In theory: B. the bulb glows dimply because there are relatively low concentrations of ions in solution.

In practice: D. an current flows because a key property of a weak acid is that it does not produce enough ions to carry the electricity through the solution.

18. The following reaction occurs:

HNO3(aq) + Ba(OH)2(aq) --> H2O(l) + Ba(NO3)2(aq)

What is the net ionic equation for this reaction?

Solution:

This is an acid (H+) base (OH-) reaction. The participating ions are H+ and OH- so:

H+ + OH- --> H2O(l)

19. Water dissolves many ionic compounds because:

Solution: Water is polar which gives it attractive forces that pull ions into solution.

20. Which of the following forms of matter has a volume that is highly dependent upon temperature and pressure?

A) solid - doesn't change size much when heated, cooled, pressed on. NOT!

B) liquid - doesn't change size much when heated, cooled,pressed on. NOT!

C) gas - Boy! does this one get compressed!!!! RIGHT ON!

D) colloid - just a really small particle.

21. Which of the following is a unit of pressure?

Solution: learn units of pressure

22. Which instrument is used to measure gas pressure?

Solution: watch the weather channel!!!!!

C. barometer

23. According to the Ideal Gas Law, pressure is directly proportional to ________.

Solution: Write the Ideal Gas Law.

PV = nRT

What can't be!

R is a constant. V is on same side so can't be directly proportional.

Only options are n and T. N isn't a choice.

24. According to the Ideal Gas Law, volume is inversely proportional to ________.

Solution: What can't be!

R is a constant. N and T are on opposite side of = so V is directly proportional to them.

Only option is P

25. An automobile tire at a pressure of 2.38 atm at 298K is heated to 360K. If the volume of the tire is constant, what is the new pressure?

Solution: Initial/Final Ideal Gas Law

```
P1V1    n1RT1
---- = -----
P2V2    n2RT2
```
Volume - constant, R - constant, n is constant since air is not moving in or out of tire.

Cancel:

```
P1      T1
--  =  --
P2      T2
```
Plug in values:

```
2.38    298
---- = ----
x     360

```
X = 2.88 atm

26. A sample of an ideal gas occupies 65.0 L at 4.45 atm and 350K. How many moles of gas are in the sample?

Solution: sue Ideal Gas Law

PV = nRT

(4.45 atm)(65.0 L) = n(0.0821 L atm /mole K)(350K)

n = 10.1 mol

27. According to the following equation, what volume of BrCL(g) at 0C and 1.0 atm is required to produce 4.4 L of Cl2 at 0C and 1.0 atm?

2BrCl(g) --> Br2(g) + Cl2(g)

2 BrCl produces 1 Br2 and Cl2

Solution:

Set up ratio.

```
2    x
- = ---     x = 8.8L
1   4.4
```
28. What volume will 128.2 g SO2(g) occupy at a pressure of 500 torr and a temperature of 127C?

This one takes knowledge and effort.

Solution:

Need molar mass of SO2. 32 + 2(16) = 64 g/mol

```
128.2 g
------- SO2(g) = 2 moles so n = 2.0
64 g/mol
```
T must be in K and K = C + 273, so K = 127 + 273 = 400 K

Pressure must be in atm so from given info on first page:

```
1 atm      1 mm Hg
500 torr x --------- x -------- = 0.657 atm
760 mm Hg    1 torr
```
Plug into Ideal Gas Law:

PV = nRT

(0.657 atm)(V) = (2 mol)(0.0821 L atm/mole K)(400 K)

V = 99.8 L

29. What is the density (g/L) of HBr(g) at 14C and 1.5 atm?

Solution: determine molar mass of HBr = 1 + 79.9 = 80.9

Convert C to K, K = 14 + 273 = 287 K

```                         PM
From front page Use D = ----
RT
Plug in values:               (1.5 atm)(80.9 g/mole)
D = ---------------------------
(0.0821 L atm/mole K)(287K)

D = 5.15 g/L = 5.2 g/l
```
30. Which if the following gases has the highest density at STP?

Solution: Since P, R, and T are constant only molar mass determines density. THe one with the greatest molar mass is the answer.

Xe @ 131.3

31. What is the partial pressure of N2(g) in a mixture of N2(g) and O2(g), if 1 mole of N2(g) and 3 moles of O2(g) share a vessel? The total pressure in the vessel is 12 atm.

Solution:

Need mole fraction so first determine total number of moles.

Xtot = 1 + 3 = 4

Next determine fraction that is N2(g)

```                1
XA  =  -
4
```
Last using partial pressure equation from front page of test.

PA = (XA)(Ptot)

PA = (1/4)(12) = 3 atm