The Hall process for the production of aluminum involves the reaction of aluminum oxide with elemental carbon to give aluminum m
etal and carbon monoxide. If the yield of this reaction is 82% and aluminum ore is 71% by mass aluminum oxide, what mass of aluminum ore must be mined in order to produce 1.0 x 103 kg (1 metric ton) of aluminum metal by the Hall process?
2 answers:
Mass of aluminium ore = 1.629.10³ kg
<h3>Further eplanation </h3>Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated
General formula:
Percent yield = (Actual yield / theoretical yield )x 100%
An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients
The Hall process:
Al₂O₃ + 3C ---> 2 Al + 3CO
To produce 1.0 x 10³ kg (1 metric ton) of aluminum metal(Al) and the yield of this reaction is 82% then
theoretical yield = An actual yield : percent yield
theoretical yield = 10³ : 0.82(82%)
theoretical yield = 1.219 .10³ kg = 1.219 .10⁶ gr mass of Al
atom mass Al = 27
mole Al = 1.219 .10⁶ : 27
mole Al = 4.5.10⁴
From reaction :
mole Al₂O₃ : mole Al = 1 : 2 , so
mole Al₂O₃ = 0.5 x mole Al
mole Al₂O₃ = 0.5 x 4.5.10⁴
mole Al₂O₃ = 2.25.10⁴
Molecular mass Al₂O₃ = 102 gr/mole
mass Al₂O₃ = mole x Molecular mass
mass Al₂O₃ = 2.25.10⁴ x 102
mass Al₂O₃ = 2.295.10⁶ gr
Because aluminum ore is 71% by mass aluminum oxide,then :
mass of aluminum ore = 0.71 (71%) x 2.295.10⁶ gr
mass of aluminum ore = 1.629.10⁶ gr = 1.629.10³ kg
<h3>Learn more </h3>The mass of one mole of raindrops
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Keywords: mole,The Hall process,mass of aluminum ore
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Answer:
ΔH = - 2020.57 kJ/mol
Explanation:
Given that :
mass of propanol = 1.685 g
the molar molar mass = 60 g/mol
Thus; the number of moles = mass/molar mass
= 1.685 g/60 g/mol
= 0.028 g/mol
However ;
ΔH = heat capacity C × Δ T
Given that:
The temperature increases from 298.00 K to 302.16 K.
Then ;
Δ T = 302.16 K - 298.00 K
Δ T = 4.16 K
heat capacity C = 13.60 kJ/K
∴
ΔH = 13.60 kJ/K × 4.16 K
ΔH = 56.576 kJ
The equation of the given reaction can be represented as :
Thus for 0.028 mol of heat liberated; ΔH = 56.576 kJ
For 1 mole of heat liberated now:
ΔH = 56.576 kJ/0.028 mol
ΔH = 2020.57 kJ/mol
SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;
ΔH = - 2020.57 kJ/mol