Question

A mixture of two solid elements with a mass of 1.52 g was treated with an excess of hydrochloric acid. A volume of 0.896 dm3 of a gas was liberated in this process and 0.56 g of a residue remained which was undissolved in the excess of the acid. In another experiment, 1.52 g of the same mixture were allowed to react with an excess of a 10 % sodium hydroxide solution. In this case 0.896 dm3 of a gas were also evolved but 0.96 g of an undissolved residue remained. In the third experiment, 1.52 g of the initial mixture were heated to a high temperature without access of the air. In this way a compound was formed which was totally soluble in hydrochloric acid and 0.448 dm3 of an unknown gas were released. All the gas obtained was introduced into a one litre closed vessel filled with oxygen. After the reaction of the unknown gas with oxygen the pressure in the vessel decreased by approximately ten times (T = const).

Write chemical equations for the above reactions and prove their correctness by calculations.

In solving the problem consider that the volumes of gases were measured at STP and round up the relative atomic masses to whole numbers.

Answer 1

(a) Reaction with hydrochloric acid: 1.52 g – 0.56 g = 0.96 g of a metal reacted and 0.896 dm3 of hydrogen (0.04 mol) were formed. 0.96

combining mass of the metal: 11.2 x 0.96/0.896 = 12 g

Possible solutions:

| Relative atomic mass of the metal | Oxidation number | Element | Satisfying? |
| 12 | I | C | No |
| 24 | II | Mg | Yes |
| 36 | III | Cl | No |

Reaction: Mg + 2 HCl → MgCl2 + H2

(b) Reaction with sodium hydroxide:

1.52 g – 0.96 g = 0.56 g of an element reacted, 0.896 dm3 (0.04 mol) of hydrogen were formed.

combining mass of the metal: 11.2 x 0.56/0.896 = 7 g

Possible solutions:

| Relative atomic mass of the element | Oxidation number | Element | Satisfying? |
| 7 | I | Li | No |
| 14 | II | N | No |
| 21 | III | Ne | No |
| 28 | IV | Si | Yes |

Reaction: Si + 2 NaOH + H2O → Na2SiO3 + 2 H2

(c) Combining of both elements:

0.96 g Mg + 0.56 g Si = 1.52 g of silicide MgxSiy

w(Mg) = 0.96 g/1.52 g = 0.63

w ( Si) = 0.56 g/1.52 g =0.37

x : y = 0.63/24 : 0.37/28 = 2:1

silicide: Mg2Si

(d) Reaction of the silicide with acid:

Mg2Si + 4 HCl → 2 MgCl2 + SiH4

n(Mg2SI) =1.52 g/76 g mol-1 = 0.02 mol

n( SiH4) = 0.448 dm3/22.4 dm3 mol-1 = 0.02 mol

(e) Reaction of silane with oxygen:

SiH4 + 2 O2 → SiO2 + 2 H2O

V = 1 dm3

On the assumption that T = const: p2 =n2/n1 p1

n1(O2) = 1dm3/22.4 dm3 mol-1 =0. 0446 mol

Consumption of oxygen in the reaction: n(O2) = 0.04 mol

The remainder of oxygen in the closed vessel:

n2(O2) = 0.0446 mol – 0.04 mol = 0.0046 mol

p2 = 0 .0046 mol/0 .0446 mol x p1 = 0.1 p1