Chem. 12 Review Test #2 Name:

Block:

Pick two formulas that match each classification:

1. b g Acid a) Ba(NO₃)₂ e) NaBr

2. c f Covalent Nonacid b) CH₃COO H f) C₃H₈

3. a e Salt c) H₂O g) H₃PO₄

4. d h Base d) Al(OH)₃ h) Ca(OH)₂

5. A student wants to determine the molarity of 100.0 mL of NaCl solution. She weighs another empty beaker and determines the mass to be 56.23g. She transfers all of the solution to this beaker and weighs it and finds the mass to be 159.99g. She proceeds to evaporate off the water until it is dry and measures the mass of the beaker and residue to be 58.69g. What is the molarity of the solution?

58.69 – 56.23 = 2.46

Molarity = 2.46 g x 1 mole

58.5 g = 0.421 M

.100 L

6. How many grams of AlCl₃ are required to prepare 250 mL of 0.200 M solution?

0.250L x 0.200 mole x 133.5 g = 6.68 g

1 L 1 mole

7. How many litres of 0.350 M MgCl₂ are needed to provide 250.0 g of MgCl₂?

250 g x 1 mole x 1 L = 7.50 L

95.3 g 0.350 mole

8. 50.6 g of AlCl₃ is dissolved in 250.0 mL H₂O, calculate [Al³⁺] and [Cl^-].

50.6 g x 1 mole

Molarity = 133.5 g = 1.516 M AlCl₃ ® Al³⁺ + 3Cl^-

0.250 L 1.516 M 1.52 M 4.55 M

9. 600 mL of 0.200 M H₂SO₄ reacts with 600 mL of 0.200 M NaOH. Calculate concentration of the excess acid in the new solution.

H₂SO₄ + 2NaOH --------> Na₂SO₄ + 2HOH

0.600L x 0.200 mole = .120 mol 0.600L x 0.200 mol = .120 mole

1 L 1 L

I 0.120 mole 0.120 mole

C - 0.060 mole - 0.120 mole

E = 0.060 mole = 0.000 mole

Beware subtraction- note the loss of 1 significant figure!

Sometimes numbers disappear during subtraction!

Total Volume = 1200 mL = 1.20 L Molarity = 0.060 mole = 0.050 M

1.20 L

10. In three runs of a titration 0.200 M NaOH was used to neutralize a 25.0 mL sample of H₂CO₃. Calculate the molarity of the acid.

0.200 M NaOH in the burette
Initial burette reading (mL)	2.05	10.56	19.09
Final burette reading (mL)	10.56	19.09	27.80
	8.51	8.53	8.71 reject

H₂CO₃ + 2NaOH → Na₂SO₄ + 2HOH

0.0250 L 0.0.00852 L

? M 0.200 M

[ H₂SO₄] = 0.00852 L NaOH x 0.200 mole x 1 mole H₂CO₃

1 L 2 mole NaOH

0.0250 L

= 0.0341 M

11. How many grams of .0200M H₂C₂O₄ are required to neutralize 250 mL of .0250M KOH?

H₂CO₃ + 2KOH → K₂SO₄ + 2HOH

0.250L KOH x 0.0250 mole x 1 mole H₂C₂O₄ x 90.0 g = 0.281 g

1 L 2 mole KOH 1 mole

12.Complete the reaction equations.

i) Formula Equation/Chemical Equation

Sr(OH)_{2 (aq)} + ZnSO_{4 (aq)} → Zn(OH)_2(s) + SrSO_4(s)

ii) Total Ionic Equation

Sr²⁺ + 2OH^- + Zn²⁺ + SO₄^2- → Zn(OH)_2(s) + SrSO_4(s)

iii) Net Ionic Equation

Sr²⁺ + 2OH^- + Zn²⁺ + SO₄^2- → Zn(OH)_2(s) + SrSO_4(s)

13. Write the complete ionic equation for the reaction of Mg _(s) and HCl _(aq).

Mg_(s) + 2HCl_(aq) → H_2(g) + MgCl_2(aq)

Mg_(s) + 2H⁺ + 2Cl^- → H_2(g) + Mg²⁺ + 2Cl^-

14. What volume of 0.100 M H₂SO₄ is needed to neutralize 25.0 mL 0.250 M NaOH solution?

H₂SO₄ + 2NaOH → Na₂SO₄ + 2HOH

? L 0.0250 L

0.100 M 0.250 M

= 0.0250 L NaOH x 0.250 mole x 1 mole H₂SO₄ x 1L = 0.0313 L

1 L 2 mole NaOH 0.100 mole

15. Calculate all ion concentrations after 200.0 mL of 0.200 M CaCl₂ is mixed with 200.0 mL of 0.300 M AlCl₃.

CaCl₂ ® Ca²⁺ + 2Cl^-

200.0 x 0.200 M = 0.100 M 0.200 M

400.0

AlCl₃ ® Al³⁺ + 3Cl^-

200.0 x 0.300 M = 0.150 M 0.450 M

400.0

[Ca²⁺] = 0.100 M

[Al³⁺] = 0.150 M

[Cl^-] = 0.200 M + 0.450 M = 0.650 M

16. An impure sample of CaC₂O₄ weighing 0.633 g is dissolved in water and the reacted with 12.20 mL of 0.120 M KMnO₄. Calculate the percent by mass of CaC₂O₄ in the original sample.

2MnO₄^- + 5C₂O₄^2- + 16H⁺ ® 2Mn²⁺ + 10CO₂ + 8H₂O

0.01220 L ? g

0.120 M

0.01220 L MnO₄^- x 0.120 moles x 5 moles C₂O₄^2- x 128.1 g CaC₂O₄ = 0.4688 g CaC₂O₄

1 L 2 mole MnO₄^- 1 mole

% = 0.4688 g pure CaC₂O₄ x 100 % = 74.1 %

0.633 g impure CaC₂O₄