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Practice Problems 8 Chapter 3 CHE 151
Graham/
1. In the reaction P 4 + 5 O 2 2 P 2 O 5
How many moles of oxygen are consumed each time 0.500 mole of P 2 O 5 is formed? 0.500 mole P 2 O 5 x 5 mole O 2 = 1.25 mole O 2 2 mole P 2 O 5
2. How many moles of oxygen are consumed in burning 125 g of benzene, C 6 H 6?
2 C 6 H 6 + 15 O 2 6 H 2 O + 12 CO 2
125 g C 6 H 6 x 1 mole C 6 H 6 x 15 mole O 2 = 12.0 mole O 2 78.41 g C 6 H 6 2 mole C 6 H 6
3. The equation for the neutralization of Citric acid is as follows:
H 3 C 6 H 5 O 7 + 3NaOH Na 3 C 6 H 5 O 7 + 3H 2 O
How many grams of sodium hydroxide are required to react with 10.50 g
of Citric acid?
10.50 g CA x 1 mole CA x 3 moles NaOH x 40.00 g NaOH = 6.558 g NaOH
192.12 g CA 1 mole CA 1 mole NaOH
4. In reacting aluminum carbonate with hydrochloric acid according to the equation
below 57.6 mg of carbon dioxide was formed.
Al 2 (CO 3 ) 3 + 6 HCl 2 AlCl 3 + 3 H 2 O + 3 CO 2 a.) What mass of aluminum carbonate was reacted? 57.6 mg CO 2 x 10-3g CO 2 x 1 mole CO 2 x 1 mole Al 2 (CO 3 ) 3 x 233.99 g Al 2 (CO 3 ) 3 = 1 mg CO 2 44.01 g CO 2 3 mole CO 2 1 mole Al 2 (CO 3 ) 3 b.) If this reaction has a 63.1% yield what mass of water will be produced? 57.6 mg CO 2 x 10-3g CO 2 x 1 mole CO 2 x 3 mole H 2 O x 18.02 g H 2 O x 63.1 g H 2 O 1 mg CO 2 44.01 g CO 2 3 mole CO 2 1 mole H 2 O 100 g H 2 O
Molar Mass C 6 H 6 78.41 g/mole Molar masses: Citric acid (CA) = 192.12 g/mole NaOH = 40.00 g/mole Molar masses: Al 2 (CO 3 ) 3 = 233.99g/mole CO 2 = 44.01 g/mole 0.102 g Al 2 (CO 3 ) 3 1.49 x 10
5. a.) What is the theoretical yield of oxygen from the decomposition of 5.00 g
of potassium chlorate?
2 KClO 3 3 O 2 + 2 KCl
5.00 g KClO 3 x 1 mole KClO 3 x 3 mole O 2 x 32.00 g O 2 = 1.96 g O 2 122.55 g KClO 3 2 mole KClO 3 1 mole O 2
b.) If the amount of oxygen actually obtained from the decomposition
is 1.50 g. Calculate the percent yield for this reaction.
% yield = Actual x 100 1.50 g x 100 = 76.5% yield O 2 Theoretical 1.96 g
6. Chlorine and fluorine react to form gaseous chlorine trifluoride, ClF 3.
You start with 3.40 moles of Cl 2 and 7.16 moles of F 2.
a.) Write a balanced equation for the reaction.
Cl2(g) + 3 F2(g) 2 ClF3(g)
b.) What is the limiting reactant?
Cl 2 : 3.40 mole Cl 2 x 2 mole ClF 3 = 6.80 mole ClF 3 1 mole Cl 2 F 2 : 7.16 mole F 2 x mole ClF 3 = 4.77 mole ClF 3 3 mole F 2
c.) What is the theoretical yield (in grams) of ClF 3?
7.16 mole F 2 x 2 mole ClF 3 x 92 45 g ClF 3 = 441 g ClF 3 3 mole F 2 1 mole ClF 3
d.) How many moles of the excess reactant remain unreacted?
7.16 mole F 2 x 1 mole Cl 2 = 2.39 mole Cl 2 consumed 3 moles F 2 3.40 mole Cl 2 - 2.39 moles Cl 2 = 1.01 mole Cl 2 remain (initial) (consumed) Molar masses: KClO 3 = 122.55 g/mole O 2 = 32.00 g/mole Thus, F 2 is the limiting reactant Molar mass: ClF 3 = 92.45 g/mole
10. What is the molarity of a KI solution that has 13.39 g KI in 450.0 mL of
solution?
13.39 g KI x 1 mole KI x 1 mL soln. = 0.1793 mole KI 0.1793 M 450.0 mL soln. 166.0 g KI 10 -3^ L soln 1 L soln
11. How many grams of NaCl (molar mass 58.44 g/mole) are present in 325.0 mL
of a 0.5055 M NaCl solution?
325.0 mL soln. x 10 -3^ L soln. x 0.5055 mole NaCl x 58.44 g NaCl = 9.601 g NaCl 1 mL soln. 1 L soln. 1 mole NaCl
12. How many liters of 3.50 M NaC 2 H 3 O 2 (molar mass 82.03 g/mole) can be
prepared from 7.50 g of the salt?
7.50 g NaC 2 H 3 O 2 x 1 mole NaC 2 H 3 O 2 x 1 L soln. = 0.0261 L. soln. 82.03 g NaC 2 H 3 O 2 3.50 mole NaC 2 H 3 O 2
13. Describe how you would prepare 1 liter of 0.15 M KI solution. KI (molar mass=
166.0 g/mole) 0.15 mole KI x 166.0 g KI = 24.9 1 L soln. 1 mole KI
14. To what total volume must 55.4 mL of 6.00 M HCl acid be diluted to produce
a 1.75 M HCl acid solution?
M 1 V 1 = M 2 V 2 V 2 = M 1 V 1
M 2
6.00 mole HCl 0.0554 L soln. V 2 = 1 L soln = 0.190 L
1.75 mole HCl
1 L soln.
15. How much solvent must be added to 250.0 mL of a 1.500 M NaCl solution to
decrease its concentration to 0.9500 M?
V 2 = M 1 V 1 = (250.0 mL)(1.500 M) = 394.7 mL (final volume total) M 2 0.9500 M
394.7 mL - 250.0 mL =
(final) (initial) Molar mass: KI = 166.0 g/mole 25 g KI dissolved in enough H 2 O to make 1 L of solution M 1 = 6.00 mole HCl 1 L soln V 1 = 55.4 mL = 0.0554 L M 2 =1.75 mole HCl 1 L soln V 2 =?
144.7 mL must be added
16. As a Lab Tech, you are asked to prepare 250.0 mL of 0.25 M KOH. You are
all out of KOH solid, but you do find a solution of 6.0 M KOH on the shelf.
Show with a calculation how you would use the available solution to fill the
request.
V 1 = (0.25 M)(250.0 mL) = 6.0 M
17. What volume (in liters) of a 4.50 M HCl solution is needed to react
completely with 0.500 L of a 6.00 M Na 3 PO 4 solution according to the
following reaction?
Na 3 PO4(aq) + 3 HCl(aq) 3 NaCl(aq) + H 3 PO4(aq)
0.500 L Na 3 PO 4 x 6.00 mole Na 3 PO 4 x 3 mole HCl x 1 L HCl soln = soln. 1 L Na 3 PO 4 1 mole Na 3 PO 4 4.50 mole HCl soln.
18. How many grams of Cr(OH) 3 can be produced from the reaction of 1.50 L of
0.275 M Cr(NO 3 ) 3 with excess 1.00 M NaOH? (molar Mass Cr(OH) 3 = 103.0 g/mole)
Cr(NO 3 )3(aq) + 3 NaOH(aq) Cr(OH)3(s) + 3 NaNO3(aq)
1.50 L Cr(NO 3 ) 3 x 0.275 mole Cr(NO 3 ) 3 x 1 mole Cr(OH) 3 x 103.0 g Cr(OH) 3 = soln. 1 L Cr(NO 3 ) 3 1 mole Cr(NO 3 ) 3 1 mole Cr(OH) 3 soln.
19. Some sulfuric acid is spilled on the lab bench. It can be neutralized by
sprinkling sodium bicarbonate on it and then mopping up the resultant
solution. The reaction is as follows:
2 NaHCO 3 + H 2 SO 4 Na 2 SO 4 + 2CO 2 + 2H 2 O
Sodium bicarbonate is added until the fizzing due to the formation of CO2(g)
stops. If 35 mL of 6.0 M H 2 SO 4 was spilled, what is the minimum mass of
NaHCO 3 that must be added to the spill to neutralize the acid?
35 mL H 2 SO 4 x 10-3L soln. x 6 mole H 2 SO 4 x 2 mole NaHCO 3 x 84.01 g NaHCO 3 = soln. 1 mL soln. 1 L soln. 1 mole H 2 SO 4 1 mole NaHCO 3 Molar mass NaHCO 3 84.01 g/mole
M 1 V 1 = M 2 V 2
V 1 = M 2 V 2
M 1
- mL of 6.0 M KOH solution diluted to exactly 250.0 mL with H 2 O 2.00 L HCl soln. 42.5 g Cr(OH) 3 35 g NaHCO 3
