Stoichiometry

Question 1

In the conical flask, there is a neutral solution of potassium sulfate which still contains the indicator used in the titration. Describe how you could obtain a solution of potassium sulfate without the indicator.

Answer 1

(add activated) charcoal/ carbon;
filter out the charcoal;

Question 2

Potassium hydrogen sulfate can be made by the following reaction.

KOH(aq) + H2SO4(aq) KHSO4(aq) + H2O(l)

Suggest how you could make a solution of potassium hydrogen sulfate without using an indicator.

Answer 2

make solution of potassium sulfate as above;

add same volume/ amount of acid again;

Question 3

Describe a test which would distinguish between aqueous solutions of potassium
sulfate and sulfuric acid

test?
result?

Answer 3

test: reactive metal/ name or formula of suitable metal, e.g. Mg/Fe/Zn;

result: bubbles or gas or hydrogen or H2 evolved/ dissolves;

Question 4

The Atacama desert in Chile has deposits of the salt sodium nitrate. Very large amounts of this salt were exported to Europe for use as a fertiliser. After the introduction of the Haber process in
1913, this trade rapidly diminished.

(a) Explain why the introduction of the Haber process reduced the demand for
sodium nitrate.

Answer 4

(Haber process makes) ammonia/NH3 ;

(ammonia converted into) fertilisers / nitrates

Question 5

(ii) Suggest why surface deposits of sodium nitrate only occur in areas with very low rainfall such as desert areas.

(iii) The desert has smaller surface deposits of potassium nitrate.
Suggest why potassium nitrate is a better fertiliser than the sodium salt.

Answer 5

ii) sodium nitrate would dissolve (in rain)/ soluble (in
water)

iii) potassium (is required by plants as well as nitrogen)

Question 6

All nitrates decompose when heated. The extent to which a nitrate decomposes is
determined by the metal in the salt.

(i) Sodium nitrate decomposes to form sodium nitrite, NaNO2.
Write the equation for decomposition of sodium nitrate.

Answer 6

2NaNO3 → 2NaNO2 + O2

Question 7

Sodium nitrite is a reducing agent.

What would be observed if an excess of sodium nitrite solution was added to a solution of acidified potassium manganate(VII)?

Answer 7

(colour changes) from pink / purple;

to colourless / decolourised;

Question 8

Copper(II) nitrate decomposes to form copper(II) oxide, nitrogen dioxide and oxygen.

What is the relationship between the extent of decomposition and the reactivity of the
metal in the nitrate?

Answer 8

the more reactive the metal the lower rate of decomposition

Question 9

The equation for the decomposition of copper(II) nitrate is given below.

2Cu(NO3)2 → 2CuO + 4NO2 + O2

(i) Predict what you would observe when copper(II) nitrate is heated.

Answer 9

(changes from) blue solid/ blue crystals;

black solid formed;

brown gas / brown vapour/(pungent) smell;

Question 10

Copper(II) nitrate forms a series of hydrates with the formula Cu(NO3)2.xH2O.

All these hydrates decompose to form copper(II) oxide.
1 mole of Cu(NO3)2.xH2O forms 1 mole of CuO.

What is meant by 1 mole of a substance?

Answer 10

Avogadro(’s) number/ constant/6.02 × 10^23;

Question 11

7.26 g of a hydrate, Cu(NO3)2.xH2O, formed 2.4 g copper(II) oxide.

i) number of moles of CuO formed =

ii) number of moles of Cu(NO3)2.xH2O in 7.26 g =

iii) mass of 1 mole of Cu(NO3)2.xH2O = ___g

iv) mass of 1 mole of Cu(NO3)2 is 188 g

the value of x in this hydrate =

Answer 11

M1
(number of moles of CuO formed = ) 0.03;

M2
(number of moles of Cu(NO3)2.xH2O in 7.26g = ) 0.03;

SAME bc both 1 mole in formula.

M3
(mass of 1 mole of Cu(NO3)2.xH2O 7.26 ÷ 0.03 =) 242(g);
(mass of 1 mole of Cu(NO3)2 is 188g)
+ mass of 1 mole - mass/moles
OR 1/0.03 = 33.3.
x 7.26 = 242.

M4
the value of x = 3;

M3 – 188 ÷ 18. 242 - 188 = 54.
2x1 + 16 = 18. Molecular mass of H2O.

54/18 = m/M = 3
x = 3 (3 moles in formula)

Question 12

Ions:

Arsenic
Silicon
Phosphide
Barium
Francium

Answer 12

As3+
Si4+
P3-
Ba2+
Fr+

Question 13

CO2 Covalent Bond

Answer 13

M1 2 double bonds, one between each O and the C atom;
M2 each O has 8 outer electrons;
M3 each C has 8 outer electrons;

Question 14

Hydrocarbon

Answer 14

compound containing carbon and hydrogen only

Question 15

General formula of straight-chain alkanes & alkenes

Answer 15

alkanes - nH2n+2

alkenes - CnH2n

(small n’s and 2’s)

Question 16

Empirical formula of X

Answer 16

1. Divide each percentage by atomic mass
2. Divide all these values by the lowest value you get
3. These are the ratios - atom shows up that many times.

Question 17

One of the samples of copper(II) oxide is impure.

Identify this sample and suggest an explanation why the percentage of copper in this sample is bigger than in the other three samples.

Answer 17

the last one;

not all the copper oxidised;

Question 18

Define the following

(i) the mole

(ii) the Avogadro constant

iii) Relative atomic mass

iv) Relative molecular mass

Answer 18

i) (the number of particles which is equal to the number of atoms in) 12g of carbon 12 OR Avogadro’s constant

ii) (Avogadro’s constant is the) number of particles / atoms / ions / molecules in one mole of
a substance

iii) the average mass of the isotopes/atoms of an element compared to 1/12th the mass of an atom of 12C (carbon-12).

iv) relative molecular mass: the sum of the relative atomic masses.

Question 19

4.8g of calcium is added to 3.6 g of water. The following reaction occurs.
Ca + 2H2O → Ca(OH)2 + H2

Which reagent is in excess? Explain your choice.

Answer 19

Ca is in excess (because 0.2 moles / 3.6g of water will react with) 0.1moles/4.0g of Ca [1]

there is more than that / there are 0.12 moles / 4.8g of Ca [1]

Question 20

Calculate the mass of the reagent named in (ii) which remained at the end of the
experiment.

Answer 20

excess x atomic mass

0.02 x 40 = 0.8

Question 21

carbon ion

Answer 21

C4-

Question 22

(c) 0.07moles of silicon reacts with 25g of bromine.

Si + 2Br2 SiBr4

(i) Which one is the limiting reagent? Explain your choice.

Answer 22

silicon is limiting reagent

0.07 moles of Si and 25/160 = 0.156 moles of Br2

because 0.14 (2 × 0.07) < 0.156