22 Solutions Stoichiometry Practice Problems with Answers PDF

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1.

Balance the following equations. Then write NET ionic equations for each. Assuming 1L of total solution, calculate the concentration of each ion left in solution after reaction. a)

K2SO4(s) +

BaCl2(aq) 

HNO3(aq) 

b) c)

Ca(OH)2(aq) +

d)

Pb(NO3)2(aq) +

e)

NaOH(aq) +

f)

Na2CO3(aq) +

Cl2(aq) +

BaSO4(s) +

NaBr(aq)  NaBr(aq) H3PO4(aq)

KCl(aq) Ca(NO3)2(aq) +

Br2(aq) +

 

H3PO4(aq) 

H2O(l)

NaCl(aq)

PbBr2(s) +

NaNO3 (aq) Na3PO4(aq) + H2O(l)

Na3PO4(aq) +

H2O(l) +

CO2(g) Pb(CH3COO)2(aq) + g) NH4CH3COO(aq) 2.

PbS (s) +

Calculate the number of: (a) (b) mol) (c) (0.05

3.

(NH4)2S(aq) 

grams in 0.25 mol of magnesium hydroxide. moles in 11.7 g of sodium chloride.

(14.5 g) (0.2

moles in 4.9 g of hydrogen sulphate. mol)

Calculate the molarity of solutions containing the following amounts of solute: (a)

31.5 g HNO3 per litre of solution.

(0.5

M) (b)

26.5 g Na2CO3 per 200 mL of solution.

(1.25

66.57 g AlCl3 per 400 mL of solution.

(1.25

M) (c) M) 4.

How many moles of solute are there in 100 mL of a 0.100 M solution? (0.01 mol)

5.

How many grams of NaCl are there in 250 mL of a 2.50 M NaCl solution? (36.6 g)

6.

What volume of a 0.75 M solution could be prepared from 500 g of Na2SO4? (4.7 L)

7.

8.

One litre of 12 M HCl is diluted to 20 L. What is the molarity of the diluted solution? (0.6 M) How many grams of potassium bromide could be obtained by evaporating to dryness 50 mL of a 0.5 M solution of the salt? (2.98 g)

9.

If 200 mL of 0.300 M Na2SO4 solution are evaporated to dryness, how many grams of sodium sulphate will be obtained? (8.52g)

1.

Calculate the mass of solute in each of the following solutions: (a) (b) g) (c) g) (d)

1.50 L of 0.050 M HNO3 500 mL of 0.060 M C12H22O11

(4.76 g) (10.3

40.0 mL of 0.970 M Ca(OH)2

(2.87

18.60 L of 1.75 M K2SO4

(5

670 g) 2.

Calculate the molarity of each of the following solutions: (a)

45 g of H2SO4 in 1.0 L of solution

M) (b) 13.0 g of NaOH in 5.0 L of solution (0.065 M) (c) 18.0 g of HCl in 75 mL of solution M) (d) 50 mg of HNO3 in 10.0 mL of solution

(0.46

(6.6

(0.0793 M) 3.

What volume of 0.10 M potassium sulphate would contain 57 g of the salt? (3.3 L)

4.

Calculate the molarity of a 500 mL volume of solution containing 2.5 g of 98.0% phosphoric acid. (0.050 M)

5.

How would you prepare 120.00 g of a 4.00% by mass BaCl2 solution, starting with BaCl2 • 2H2O and water? (5.63 g BaCl2 • 2H2O + 114.37 g H2O)

6.

How many grams of 6.0% by mass KOH solution will yield 150 g of KOH? (2 500 g)

7.

How many grams of copper (II) sulphate pentahydrate are required to prepare 250 mL of a solution that is 1.25 M? (78.1 g)

8.

How much sulphuric acid (95.0% by mass) is needed in the preparation of 1 000 g of a 25.0% by mass solution of the acid? (263 g)

9.

Express the concentrations of the following solutions as molarity: (a) 95% solution of H2SO4, density 1.79 g/mL M) (b)

(17.3

35 g of NaOH dissolved in 150 g water, density 1.57 g/mL (7.43 M) (c) 38% solution of HCl, density 1.43 g/mL (14.9 M) (d) 60 g of KOH dissolved in 300 g water, density 1.30 g/mL (3.87 M) (e) 28.3 g of KOH dissolved in enough water to make 2L. (0.253 M) (f) 1.0 g of NaOH dissolved in enough water to make 1 000 mL. (0.025 M) (g) 10% by mass hydrochloric solution; the density is 1.047 g/mL (2.87 M)

1.

515 g of copper metal is reacted with 4.8 L of a 3.7 M solution of HNO3 (aq) according to the following equation. : 3Cu (s)+8HNO3 (aq) → 4H2O (g)+3Cu(NO3)2 (aq) + 2NO (g) Calculate how much of the copper metal remains after the reaction is complete. [ 91.8 g ]

2.

A 105 mL sample of aluminum nitrate solution is completely reacted with excess potassium hydroxide solution producing 2.79 g Al(OH)3 (s) precipitate. a) b)

Write the ionic equation and the net ionic equation for the reaction. Determine the concentration of the Al(NO3)3(aq) used. [ 0.340 M ]

3.

In an acid - base titration 25.00 mL of a NaOH solution were completely neutralized by 33.34 mL of 0.0970 mol/L hydrochloric acid . a) Write the balanced equation for this reaction. b) Calculate the concentration of the NaOH solution. [ 0.129 M ]

4.

What mass of sodium chloride is necessary to precipitate 4.94 g of silver chloride from excess of silver nitrate solution? [1.99 g]

5.

Iodine reacts with a nitric acid solution according to the following chemical equation : 10HNO3 (aq) + 1I2 (s) → 2HIO3 (aq) + 10NO2(g) + 4H2O (l) If 17.5 g of iodine is reacted with 2.4 L of 0.41 M nitric acid solution then determine: a) the limiting reagent. [ iodine] b) how much iodic acid, in grams, that should theoretically be produced . [ 24.25 g ] c) the percentage yield If 11.6 g of iodic acid is actually produced. [ 47.8%]

6.

35 mL of 1.1 M sodium phosphate [Na3PO4] solution is mixed with 20 mL of 3.5 M calcium chloride [CaCl2] solution. A fine white precipitate is formed. a) Write the balanced chemical equation for this reaction. b) Write the Ionic equation for this reaction. c) Write the Net Ionic equation for this reaction. d) Name the precipitate which is formed.(THINK!) e) How much precipitate will be produced in grams. [ 6.0 g ] f) Name the ions that will be left in solution after the reaction. g) Calculate the concentration of each ion left in solution (recall total volume)

7.

The chemical process for making photographic film involves the chemical conversion of metallic silver to silver nitrate by the following reaction. 4HNO3 (aq) + 3Ag (s) → NO (g) + 3AgNO3 (aq) + 2H2O (l) In one industrial operation a block of silver with the dimensions 20.0 cm x 15.0 cm x 30 cm was dissolved in concentrated nitric acid. [ the density of silver is 10.5 ] a) What is the theoretical yield of silver nitrate in kilograms. [ 149 kg] b) How many litres of 7.0 M HNO3 are needed, in theory, for complete reaction to occur? [ 167 L ]

8.

If 17.4 mL of 3.50 M NaOH reacts with an excess of H2SO4, how many grams of Na2SO4 are produced? NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + H2O (l) [ 4.32 g ]

9.

Water is added to 100 mL of a 8.0 M sodium iodide solution to make 500 mL. 160 g of lead (II) nitrate is added to the solution. Calculate i) the mass of the yellow precipitate formed. [184 g] ii) the mass of any reactant left unreacted. [ 26.5 g] ii) the concentration of all ions left in solution.

1.

Calculate the pH of: a) 0.050 M H O+ b) 0.12 M HCl 3 e) 0.1 M H O+ f) 0.01 M H O+ 3

2.

3

c) 2.4 M H3O+ d) 1.0 M H2SO4 g) 1.0 x 10-3 M HCl

For each of the following, state whether the solution is acidic, basic or neutral and calculate the [H+] of the solution if it has: a) pH = 3.50 b) pH = 12.56 c) pH = 7 d) pH= 6 e) pH = 0 f) pH = - 0.5 g) pH = 9.2

3. a) Write the balanced chemical equation for the neutralization of hydrochloric acid with sodium hydroxide. b) If 35.4 mL of 0.50 M NaOH neutralize 70.3 mL of HCl, what is the concentration of the acid? 4

a) Write the balanced chemical equation for the neutralization reaction between sulphuric acid and potassium hydroxide. b) How many mL of 0.15 M potassium hydroxide are needed to neutralize 50 mL of 0.12 M sulphuric acid?

5. Write the reaction equation and calculate the volume of 0.14 M

hydrochloric acid needed to neutralize 40.3 mL of 0.17 M calcium hydroxide? 6. What is the concentration of sulphuric acid if 335 mL of it react completely with 266 mL of 0.015 M barium hydroxide? 7. The main acid component of citrus fruits such as lemons is citric acid, H3C6H5O7. Calculate the concentration of citric acid in a solution if a 15.00 mL sample is neutralized by 7.51 mL of 0.0100 M KOH. The equation for this reaction is: H3C6H5O7 + 3 KOH  K3C6H5O7 + 3 H2O 8. What volume of 0.20 M nitric acid is needed to react completely with 37 g of calcium hydroxide? 3.

In an acid - base titration 25.00 mL of a NaOH solution were completely neutralized by 33.34 mL of 0.0970 mol/L hydrochloric acid . a) Write the balanced equation for this reaction. b) Calculate the concentration of the NaOH solution.

Answers: 1. a)1.30, b) 0.9, c) - 0.38, d) 0, e) 1, f) 2, g) 3 10-4 M,

2. a) acid, 3.2 x

b) base,2.8 x 10-13 M, c) neutral, 1.0 x 10-7 M, d) acid,1.0 x 10-6 M, e) acid,1 M, f) conc. acid, 3.1 M, g) base, 6.3 x 10-10 M 3. HCl + NaOH  NaCl + H2O, [HCl]=0.25 M 4. : H2SO4 + 2 KOH  K2SO4 + 2 H2O, VKOH= 80 mL 5. 2 HCl + Ca(OH)2  CaCl2 + 2 H2O, VHCl= 98 mL. 6. H2SO4 + Ba(OH)2  BaSO4 + 2 H2O, C = 0.0119 M 7. C = 0.0167 M 8. 5.0 L HNO3 solution 9. 0.129 M...