[derbygirl2005]

1. If 15.5 g of aluminum are reacted with 46.7 g of chlorine gas, then aluminum choride, AlCl3 is formed.
a) which reactant is in excess
b) calculate the number of grams of excess
c)calculate the mass of aluminum chloride produced.

2. In a car battery, lead, lead (IV) oxide, PbO2, and sulfuric acid, H2So4, are reacted to produce lead (II) sulfate, PbSO4, and water.
A)write the balanced equation for the reaction
b) when 10.45 g of lead, 15.66 g of lead (IV) oxide, and 25.55 g of sulfuric acid are mixed, which is the limiting reactant?
c) how many grams of lead (II) sulfate will be produced?

~~~~I need help with on how to do these two problems. if someone explains it,maye i will be able to get the whole chapter. ~~~~~
again gracious

[Donaldson Tan]

The idea behind stoichiometry is that an amount of substance would react with another substance according to a certain ratio call the molar ratio.

eg. 2H₂ + O₂ => 2H₂O
2 moles of Hydrogen gas reacts with 1 mole of oxygen gas to produce 2 moles of water. The molar ratio of oxygen to hydrogen is therefore 2:1.

With reference to (1)
2Al + 3Cl2 -> 2AlC₃
the molar ratio of chlorine to aluminum is thus 3:2. Every 2 mole of Al reacts with 3 mole of Chlorine gas to produce Aluminium Chloride.

amount of Aluminum
= 15.5g
= (15.5/27) moles
= 0.574 moles

amount of Cl₂
= 46.7g
= 46.7 / 71 moles
= 0.658moles

amount of Al would react with all Cl₂
= 2/3 X 0.658
= 0.438moles

Since all Cl₂ present require less than 15.5g Al, Aluminium therefore must be in excess.

Grams of Al in excess
= (0.574-0.438)X27
= 3.66g

By Law of Mass Combination,
Mass of Aluminium Chloride produced
= 15.5 + 46.7 -  3.66
= 58.54
= 58.5g (3s.f)

Question (2) can be solved in a similar manner.