quantitive chem

Question 1

Law of conservation of mass

Answer 1

The law of conservation of mass states that no atoms are lost or made during a chemical reaction so the mass of the products equals the mass of the reactants.

Question 2

Balanced equation of magnesium and hydrochloric acid

Answer 2

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)

Question 3

Relative atomic mass (RAM)

Answer 3

Average mass of atoms in an element taking into account masses and abundance of its isotopes

Question 4

Relative formula mass (RFM)

Answer 4

Sum of RAM’s of all atoms in the formula.

Question 5

Relative formula mass of CaF2

Answer 5

CaF2 - (Ar values: Ca = 40

Question 6

Relative formula mass of C6H12O6

Answer 6

(Ar values: C = 12

Question 7

Escape of gases in reaction

Answer 7

They are both gases.

Question 8

Mean mass of magnesium produced

Answer 8

(3.3 + 3.5 + 3.2) / 3 = 3.3

Question 9

Increasing precision of results

Answer 9

Measure to more decimal places or use a more sensitive balance/apparatus.

Question 10

Avogadro’s constant

Answer 10

The number of atoms

Question 11

Formula linking mass

Answer 11

molecular mass

Question 12

Mass of 20 moles of calcium carbonate (CaCO3)

Answer 12

Mr = 100; 100 x 20 = 2000 g.

Question 13

Calculate moles of carbon dioxide from mass

Answer 13

Calculate the amount of carbon dioxide in moles in 0.32 g of carbon dioxide.

Question 14

Relative atomic mass (Ar)

Answer 14

Carbon = 12

Question 15

Moles formula

Answer 15

Moles = Mass / Mr

Question 16

Moles of ammonia from mass

Answer 16

6800000 / 17 = 400000 moles of ammonia

Question 17

Balanced equation for ammonia formation

Answer 17

N2(g) + 3 H2(g) ⇌ 2 NH3(g)

Question 18

Mass of nitrogen for ammonia

Answer 18

The mass of nitrogen needed to form 6.8 tonnes of ammonia is 5.6 tonnes.

Question 19

Moles of nitrogen from ammonia

Answer 19

The number of moles of nitrogen is 400000/2 = 200000.

Question 20

Mr of nitrogen gas (N2)

Answer 20

Mr of N2 is 28.

Question 21

Limiting reactant definition

Answer 21

The reactant that is completely used up is called the limiting reactant because it limits the amount of products.

Question 22

Hydrogen peroxide decomposition

Answer 22

Hydrogen peroxide decomposes in water to form water and oxygen.

Question 23

Balanced equation for hydrogen peroxide

Answer 23

2 H2O2(l) → 2 H2O + O2(g)

Question 24

Moles of hydrogen peroxide

Answer 24

Number of moles in 40.8 g: 40.8/34 = 1.2 moles.

Question 25

Moles of oxygen from hydrogen peroxide

Answer 25

Number of moles of O2 = 0.6 moles.

Question 26

Mass of oxygen from moles

Answer 26

Mass of oxygen = 0.6 x 32 (Mr of O2) = 19.2 g.

Question 27

Concentration formula (mass)

Answer 27

Concentration (g per dm3) = Mass (g)/Volume (dm3)

Question 28

Concentration formula (moles)

Answer 28

Concentration (mol per dm3) = nr of moles/volume (dm3)

Question 29

Potassium hydroxide neutralization

Answer 29

31.0 cm3 of potassium hydroxide solution neutralised 25.0 cm3 of 2.0 moldm−3 nitric acid.

Question 30

HNO3 + KOH → KNO3 + H2O

Answer 30

Chemical equation representing the reaction between nitric acid and potassium hydroxide.

Question 31

Concentration of potassium hydroxide solution

Answer 31

Calculated in moldm−3.

Question 32

Moles of HNO3 used

Answer 32

Calculated as Concentration x volume; 2 x 0.025 dm3 = 0.05 moles.

Question 33

Moles of KOH

Answer 33

Calculated as 0.05 moles due to a 1:1 ratio with HNO3.

Question 34

Concentration of KOH

Answer 34

Calculated as Volume = Moles/concentration; 0.05 / 0.031 = 1.61.

Question 35

Molar volume of a gas at room temperature and pressure

Answer 35

1 mole of a gas occupies 24 dm3.

Question 36

Titration

Answer 36

A technique for finding the concentration of a solution by reacting a known volume of this solution with a solution of known concentration.

Question 37

Conducting a titration

Answer 37

Steps include rinsing the pipette

Question 38

Reasons for not obtaining theoretical product amount

Answer 38

Reactions may not go to completion

Question 39

Percentage yield of a product

Answer 39

% Yield = (Actual mass of a product / Maximum theoretical mass of product) x 100%.

Question 40

Calculating percentage yield of NH3

Answer 40

If 40.5 g NH3 is produced from 20.0 mol H2

Question 41

Theoretical amount of NH3

Answer 41

Calculated as 13.3 moles from 20/1.5 based on the ratio of H2 to NH3.

Question 42

Mass of NH3

Answer 42

Calculated as 13.3 X 17 (Mr of NH3) = 227.

Question 43

Percentage yield of NH3 calculation

Answer 43

Calculated as (40.5/227) x 100 = 17.8%.

Question 44

Atom economy

Answer 44

A measure of the amount of starting materials that end up as useful products

Question 45

Reaction I for CuCl2 production

Answer 45

CuCO3(s) + 2 HCl(aq) → CuCl2(aq) + H2O(l) + CO2(g).

Question 46

Reaction II for CuCl2 production

Answer 46

CuO(s) + 2 HCl(aq) → CuCl2(aq) + H2O(l).

Question 47

Reactive formula masses

Answer 47

CuO = 79.5; HCl = 36.5; CuCl2 = 134.5; H2O = 18.

Question 48

Better atom economy reaction

Answer 48

Reaction II has a better atom economy with a total formula mass of reactants = 152.5 and formula mass of CuCl2 = 134.5