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%
