ch 7
2 Al(s) + 3 CuO(s) → Al2O3(s) + 3 Cu(s).
You react 12.0 g Al with excess CuO. How many grams of Cu form?
42.4 g Cu
12.0 g (1 mol/26.9815 g) = 0.44475 mol Al
0.44475 mol Al × (3mol CuO/2 mol Al) = 0.66712 mol CuO
0.66712 mol CuO × 63.546 g/mol CuO
Select all statements that are consistent with the Kinetic Molecular Theory of gases.
A) Gas particles are in constant, random motion and their collisions with container walls cause pressure.
B) Gas particles occupy a significant fraction of the container’s volume.
C) Gas particles exert attractive and repulsive forces on each other under normal conditions.
D) The collisions between gas molecules are perfectly elastic (no energy is lost).
A, D
Which compound below becomes a strong electrolyte when dissolved in water?
A) CO₂
B) C₆H₁₂O₆ (glucose)
C) CH₃OH (methanol)
D) Na₂SO₄
E) Br₂
D) Na₂SO₄ because it completely dissociates into ions when dissolved in water
For the reaction
Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)
describe how each change affects the rate of hydrogen gas production:
A. Increasing the concentration of HCl
B. Cooling the reaction mixture
A. Increase: higher [HCl] → more collisions between H⁺ ions and zinc atoms
B. Decrease: lower temperature → fewer particles with sufficient kinetic energy to react
2H2(g)+O2(g)→2H2O(g)
If you react 5.00 g H₂ with 25.0 g O₂,
(a) identify the limiting reactant, and
(b) Calculate the moles of H2O produced
limiting reactant O2; 1.56 mol H2O
H2: 5.00 g (mol/2.016 g) (2 mol H2O/2 mol H2)= 2.480 mol H2O
O2: 25 g (mol/32 g) (2 mol H2O/1 mol O2)= 1.56 mol H2O
A 2.50 L sample of gas at 300.0 K and 1.20 atm is heated to 450.0 K while allowed to expand to 4.00 L. What is the final pressure?
1.13 atm
P2=P1V1T2/V2T1
(1.20× 2.50x 450)/(4.00×300)
Based on the solubility guidelines, which of the following cations would form an insoluble compound with carbonate (CO₃²⁻) but not with nitrate (NO₃⁻)?
A) Na⁺
B) NH₄⁺
C) Ca²⁺
D) K⁺
E) Li⁺
C) Ca2+
For the reaction
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
predict how each of the following changes would affect the rate of reaction (increase, decrease, or no effect). Explain briefly.
A. Lowering the temperature from 500 °C to 300 °C
B. Increasing the pressure of O₂
C. Adding a platinum catalyst
A. Decrease: lower temperature → fewer particles exceed activation energy.
B. Increase: higher O₂ pressure → more frequent collisions.
C. Increase: catalyst lowers activation energy, speeding up both forward and reverse rates.
For 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s), you have 10.0 g Fe and 8.00 g O2.
(a) Identify the limiting reactant.
(b) Calculate the grams of Fe2O3 produced
Fe is limiting producing 14.30 g Fe2O3
Fe: 10 g Fe (55.845 mol/g) (2 mol Fe2O3/4 mol Fe) = 0.08953 mol Fe2O3
O2: 8 g O2 (32 mol/g) (2 mol Fe2O3/3 mol O2) = 0.16667 mol Fe2O3
0.08953 mol Fe2O3 (159.69 g/mol)
A gas mixture contains N₂ and CO₂. The partial pressure of N₂ is 625 torr, and the partial pressure of CO₂ is 0.715 atm.What is the total pressure of the mixture (in atm)?
1.54 atm
625 torr× (1atm/760 torr) = 0.822 atm
0.822+0.715=1.537 atm
What volume (mL) of a 12.0 M HCl solution must be diluted to prepare 250. mL of a 2.00 M HCl solution?
41.7 mL
M1V1=M2V2
(2.00 x 250)/12
A reaction's reactant have a potential energy of 120 kJ·mol⁻¹ and the products have a potential energy of 70 kJ·mol⁻¹. (hint: drawing the energy diagram can help visualize this)
What is the ΔH (heat of reaction) and is it exotherimic or endothermic?
-50 kJ/mol (exothermic)
ΔH=Eproducts−Ereactants
70-120
A 5.00 L container at 25.0 °C holds a gas at 0.250 atm and that contained gas has mass = 3.50 g. What is the molar mass of the gas?
68.50 g/mol
25.0 °C = 298.15 K.
n = PV/(RT) = (0.250 × 5.00) / (0.08206 × 298.15) = 0.05109 mol
3.50 g/0.05109 mol = 68.50 g/mol
You mix 150.0 mL of 0.500 M NaCl with 250.0 mL of 0.200 M NaCl. What is the final molarity of NaCl in the mixture?
0.3125 M
0.150 L × 0.500 M = 0.0750 mol
0.250 L × 0.200 M = 0.0500 mol
Total moles = 0.1250 mol
total volume = 0.150 + 0.250 = 0.400 L
0.1250 mol/ 0.400 L
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g). You have 10.0 g C3H8 and 40.0 L O2. Assume O₂ is measured at STP. Which reactant limits the reaction, and what mass of CO2 is produced?
limiting reactant: C3H8 produces 29.94 g CO2
10 g C3H8 (mol C3H8/44.095 g) (3 mol CO2/ 1mol C3H8) = 0.68035 mol CO2
40 L O2 (mol O2/22.4 L) (3 mol CO2/ 5 mol O2) = 1.07076 mol CO2
0.68035 mol CO2 (44.0095 g/mol CO2)
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
25.0 g KClO3 is fully decomposed. What volume (in L) of O₂ is produced at T = 350.0 K and P = 0.950 atm?
9.25 L O₂
25 g KClO3 (mol/122.55 g) (3 mol O₂/ 2 mol KClO3) = 0.30600 mol O2
V=nRT/P
(0.30600×0.08206×350.0)/0.950
AlCl₃(aq) + 3 AgNO₃(aq) → 3 AgCl(s) + Al(NO₃)₃(aq)
How many grams of AgCl are formed when 30.0 mL of 0.250 M AgNO₃ reacts with excess AlCl₃?
1.07 g AgCl
0.0300 L (0.250 M) = 0.0075 mol AgNO3
0.0075 mol AgNO3 (1 mol AgCl/ 1 mol AgNO3)
0.0075 mol AgCl (143.32 g/mol)