A) Aluminum reacts with chlorine gas to form aluminum chloride via the following

Question

A) Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)2AlCl3(s) What is the maximum mass of aluminum chloride that can be formed when reacting 34.0 g of aluminum with 39.0 g of chlorine?

B) C8H18(g)+O2(g)CO2(g)+H2O(g)

How many moles of water are produced in this reaction? After the reaction, how much octane is left?

C) If 49.0 % of the SO2 could be removed by reaction with powdered calcium oxide, CaO, via the reaction

SO2(g)+CaO(s)CaSO3(s)

how many tons of calcium sulfite, CaSO3, would be produced?

Explanation / Answer

A)2Al(s)+3Cl2(g)2AlCl3(s)

Aluminum: n = 34.0 / 26.98 = 1.26 moles of Al
Chlorine: n = 39.0 / (35.453 x 2) = 0.55 moles of Cl2

Cl2 is the limiting reagent so this will determine the amount of aluminum chloride that will be formed.
3 moles of chlorine gas makes 2 moles of AlCl3 so:
Moles of AlCl3 formed = 2/3 x .55moles = 0.3666 moles
Using n = m/M
0.3666 = m / (26.98 + 35.453 x 3)
m = 48.89 grams

2) we need quantative data .

3)SO2(g)+CaO(s)CaSO3(s)

SO2 and CaCO3 are equimolar.

remaining = 100-49 = 51%

produced CaSO3 = 51%of total So2

So 48.9 grams of AlCl3 will be formed

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