Question

Q7. Aluminum is extracted from the ore bauxite, which is impure aluminum oxide. 1 tonne (1000 kg) of the ore was found to have this composition: 1. aluminum oxide 825 kg 2.iron(III) oxide 100 kg 3.sand 75 kg
(a): What percentage of this ore is impurities?
(b): 1 tonne of the ore gives 437 kg of aluminum.
(i): How much aluminum will be obtained from 5 tonnes of the ore?
(ii): What mass of sand is in this 5 tonnes?
(c): What will the percentage of aluminum oxide in the ore be, if all the iron (III) oxide is removed, leaving only the aluminum oxide and sand?

Answer 1

Answer:

A) 17.5 %

B) 1) 2135 kg

2) 375 kg

C) 91.8%

Explanation:

Part (A):

1 tonne = 1000 kg = 100%

Aluminium oxide = 825 kg = 82.5%

Iron (III) oxide = 100 kg = 10%

Sand = 75 kg = 7.5%

That makes total of 100%

So, the percentage of impurities = Iron (III) oxide + Sand

=> 10 + 7.5 = 17.5%

Part (B):

(1) 1 tonne = 427 kg

Multiplying both sides by 5

=> 5 tonne = 427 * 5

=> 5 tonnes = 2135 kg

(2) 1 tonne contains 75 kg sand as impurity

=> 1 tonne = 75 kg

Multiplying both sides by 5

=> 5 tonnes = 75*5

=> 5 tonnes = 375 kg

Part (C):

If all of the iron oxide was removed, then only aluminium and sand are left.

Aluminium = 825 kg

Sand = 74 kg

Total of them makes it:

=> 825 + 74 = 899

So,

%age of Aluminium = $\frac{825}{899} * 100$

%age of Aluminium = 0.918 * 100%

%age of Aluminium = 91.8%

Answer 2

Answer:

(a) Total impurities = 17.5%

(b)

(i) 5 tons of ore yields 2185 kg of aluminium

(ii) 5 tons of ore contains 375 kg of sand

(c) Percentage of Al2O3 in purified ore = 825/900 = 91.7%

Explanation:

Q7. Aluminum is extracted from the ore bauxite, which is impure aluminum oxide.

1 tonne (1000 kg) of the ore was found to have this composition:

1. aluminum oxide 825 kg  (82.5%)

2.iron(III) oxide 100 kg  (10%)

3.sand 75 kg (7.5%)

(a) Since we are interested only in extracted aluminium, everything other than aluminium oxide is considered impurity, which includes Fe2O3 (10%) and sand (7.5%).  Total impurities = 10+7.5 = 17.5%

(b)  "1 tonne of the ore gives 437 kg of aluminum."

(i) 1 ton of ore : 437 kg Al  = 5 tons of ore : x kg of Al

Cross multiply

x = 5 tons * 437 kg / 1 ton = 2185 kg of aluminium

(ii) 1 ton of ore contains 75 kg of sand

5 tons of or contains x kg

x = 5*75/1 = 375 kg of sand

(c) If the ore is purified from Fe2O3, then for each ton of ore, there will be left

825 kg of aluminium oxide,

75 kg of sand

for a total of 900 kg.

Percentage of Al2O3 = 825/900 = 91.7%