**How many moles of oxygen molecules would be produced from the decomposition of 2.50 moles of potassium trioxochlorate (V)?**

a. 2.50 b. 3.50 c. 3.75 d. 7.50

SOLUTION

2KClO3(s) → 2KCl(s) + 3O2(g)

If 2moles of KCLO3 gives 3 moles of oxygen (O2)

2.5 moles will give Xmoles

Xmoles = 2.5 × 3/2

= 3.75 mole- :
**Na2CO3 + 2HCL 2NaCL + H2O + CO2(g). Using the above equation, what volume of carbondioxide, measured at s.t.p is liberated when 53g of sodium carbonate is dissolved in hydrochloric acid? ( 1 mole of gas occupies 22.4dm3 at s.t.p) (Na = 23, C = 12, O = 16)**

a. 44.8 dm3 b. 11.2 dm3 c. 100.1 d. 3.0 dm3 e. 22.4 dm3

SOLUTION

Na2CO3 + 2HCL → 2NaCL + H2O + CO2(g).

Molar mass of Na2CO3 = 23×2 + 12 + 16×3

46+12+48 = 106gmol–¹

No. Of moles = mass/molar mass

53/106 = 0.5moles ( Na2CO3 + 2HCL gives one mole of CO2)

No. Of moles = volume at s.t.p/ molar volume;

Volume at S.t.p = 0.5 × 22.4

11.2 dm³. [ B ] **Given that 32.0g sulphur contains 6.02 × 1023 sulphur atoms. How many atoms are there in 2.70g of alluminium? (AL = 27, S = 32)**

a. 6.02× 1023 b. 3.01× 1022 c. 6.02 × 1022 d. 5.08× 1022 3.01× 1022

SOLUTION

No. Of moles of aluminum = 2.7/27; = 0.1 moles

No. Of atoms = No. Of moles × Avogadro’s constant

0.1 × 6.02 × 10²³. = 6.02 × 10²². [ C]**If 10cm3 of distilled water is added to 10cm3 of an aqueous salt solution, the concentration of the solution**

a. increases b. decreases c. remains constant d. doubles

SOLUTION

Addition of water to a solution decreases the level of concentration of the solution, this is called a diluted solution. NOTE: if you did not add water or any other solvent and you take a particular volume out of the solution, the concentration remains the same (imagine you have a cup full of water and you dissolved a tea spoon full of sugar on it. If you take a quantity out of the cup, does the solution taste any difference? No!). Changing the amount of solute or solvent is what affects concentration of a solution. ANS:B**Which of the following quantities is the same for one mole of Br2(l) and one mole of He(e)?**

a. number of molecules b. number of atoms c. total mass of the atoms d. volume occupied at s.t.p

SOLUTION

Their number of molecules will be the same for one mole. They both contain 6.02 × 10²³ molecules. But their number of atoms is different because Br2 is diatomic (2 × 6.02 × 10²³) while He is monoatomic (6.02×10²³). Since they have different atomic masses, their total mass cannot be the same. They could have occupied the same volume since they have same number of molecules, but Br2 is a liquid, the rule only applies to gases.

ANS: [ A ]**During a titration experiment, 0.05 moles of carbon (IV) oxide is liberated. What is the volume of CO2 gas liberated?**

a. 22.40 dm3 b. 11.20 dm3 c. 2.24 dm3 d. 1.12 dm3

SOLUTION

Volume of CO2 = 22.4 × 0.05 ; 1.12dm³ [D]**If a solution contains 4.9g of tetraoxosulphate(vi) acid, calculate the amount of copper (II) oxide that will react with it [Cu = 64, O = 16, S = 32, H = 1]**

a. 0.8g b. 4.0g c. 40.0g d. 80.0g

SOLUTION

Molar mass of H2SO4 = 1×2 + 32 + 16×4; = 2+32+64

= 98mol–¹

First let’s find the number of moles of tetraoxosulphate(vi) acid,

No. Of moles = mass/molar mass

4.9/98 = 0.05moles

Molar mass of CuO = 64 + 16; = 96gmol–¹

Mass = 0.05×80; = 4.0g**Calculate the volume of carbon (IV) oxide measured at s.t.p, produced when 1kg of potassium hydrogen trioxocarbonate (IV) is totally decomposed by heat? (G.M.V at s.t.p = 22.4 dm3 , K = 39, O = 16, C = 12, H = 1)**

a. 28 dm3 b. 56 dm3 c. 112 dm3 d. 196 dm3

SOLUTION

2KHCO3(s) → K2CO3 (s) + H2O(g) + CO2(g)

Volume of CO2 = ?

Molar mass of KHCO3 = 39+1+12+16×3; = 100gmol–¹, mass = 1000g

No. Of moles of potassium hydrogen trioxocarbonate (IV), KHCO3 = 1000/100

= 10moles

The mole ratio of KHCO3 to CO2 = 2:1

Therefore, No. of moles = 10/2 = 5moles

Volume of CO2 = No. of moles × molar volume

5 × 22.4 = 112dm³**If 200cm3 of a gas at s.t.p has a mass of 0.268g, what is its molar mass? (Molar volume of a gas at s.t.p = 22.4 dm3 )**

a.300gmol-1 b 200 gmol-1 c. 150 gmol-1 d. 30 gmol-1 e. 15 gmol-1

SOLUTION

No. Of moles = volume/molar volume ; = 200/22.4 ÷ 1000 ( converting cm to dm)

8.923/1000 = 0.008923moles

Molar mass = mass/No. of moles; = 0.268/0.008923

= 30gmol–¹**What volume of oxygen at s.t.p would react with carbon to form 4.40g of CO2 according to the following equation? C(s) + O2(g) CO2(g) [ o = 16, c = 12; 1 mole of a gas occupies 22.4dm3 at s.t.p]**

a. 0.224 dm3 b. 2.24 dm3 c. 4.40 dm3 d. 4.48 dm3 e. 22.4 dm3

SOLUTION

CO2 → O2 + C

C(s) + O2(g) → CO2(g)

The mole ratio = 1:1

Molar mass of CO2 = 12 + 16×2; = 44gmol–¹

No. of moles of CO2 = mass/molar mass; = 4.40/44; = 0.1moles

Volume of oxygen = molar volume × No. of moles

22.4×0.1 = 2.24dm³. [B]**If 1 mole of sodium contains 6 × 1023 atoms, how many atom are contained in 0.6g of sodium? [Na = 23, ]**

a. 1.56 × 1023 b. 1.56 × 1022 c. 3.6 × 1023 d. 3.6× 1022

SOLUTION

No. of atoms = No. of moles × Avogadro’s constant

No. of moles = 0.6/23; = 0.026moles

0.026×6.02×10²³. = 1.56 × 10²² [B]**If 20cm3 of distilled water is added to 80cm3 of 0.50 moldm-3 hydrochloric acid, the new concentration of the acid will be**

a. 0.10 moldm-3 b. 0.20 moldm-3 c. 0.40 moldm-3 d. 2.00 moldm-3

SOLUTION

2HCl +2H2O → H2O2 + 2H2 + Cl2

VA = 20cm³, Ca = ?, Vb = 80cm³, Cb = 0.50 moldm–³, Na/Nb =

CaVa/CbVb = Na/Nb

20×Ca/0.50×80 = 1/1

Ca = 40/20; = 2.00moldm–³**Consider the reaction represented by the equation. 2Na2HCO3(s) 2Na2CO3(s) + CO2(g) + H2O(g). What volume of carbon (IV) oxide at s.t.p is evolved when 0.5 moles of Na2HCO3 is heated? [molar volume = 22.4dm3 at s.t.p]**

a. 1.12 dm3 b. 2.24 dm3 c. 5.6 dm3 d. 56.0 dm3

SOLUTION

2Na2HCO3(s) → 2Na2CO3(s) + CO2(g) + H2O(g).

2 moles of Na2HCO3 gives 1 mole of CO2; therefore 0.5/2 = 0.25moles

Volume = 0.25 × 22.5; = 5.6dm³**An iron ore is known to contain 70.0% Fe2O3 . The iron obtained from 80kg of the ore is? (Fe = 56, O = 16]**

a. 35.0kg b. 39.2kg c. 70.0kg d. 78.4kg

SOLUTION

Ore is defined as a mineral that contains a metal of interest. We will talk more about ores when we get to metals and non-metals. Firstly, let is know the amount of Fe2O3 in the ore;

%Fe2O3 = mass of Fe2O3 /mass of ore × 100

70 = mass of Fe2O2/80 × 100

Mass of Fe2O3 = 56kg

The decomposition of Fe2O3 gives

2Fe2O3→ 4Fe + 3O2

Molar mass of Fe2O3 = 56×2 + 16×3; = 160gmol–¹

Number of mole of Fe2O3 = mass/molar mass; = 56×10³/160

= 350 moles

The mole ratio of Fe2O3 to Fe is 2:4 which gives 1:2

Number of mole of Fe = 350×2; = 700moles

Mass of Fe = No. Of moles × molar mass

700×56 = 39200; = 39.2kg**In two separate experiments, 0.36g and 0.71g of chlorine combined with a metal X to give Y and Z respectively. An analysis showed that Y and Z contain 0.20g and 0.40g of X respectively. The data above represents the law of?**

A. Multiple proportion. B. Conservation of mass. C. Constant composition.

D. Reciprocal proportion

SOLUTION

The ratio of chlorine to that of X (1:2) in the two compounds Y and Z remains the same. This obeys the law of constant composition or definite proportion.

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