Question

Silver chloride and aluminum chlorate react to form silver chlorate and aluminum chloride.

a. How many grams of aluminum chlorate are required to produce 5.00g aluminum chloride?

b. How many moles of silver chlorate are produced from 1.00 moles of aluminum chlorate?

Answer 1

The balanced reaction between Silver chloride (AgCl) and aluminum chlorate {Al(ClO₃)₃} can be written as follows.

3AgCl + Al(ClO₃)₃ 3AgClO₃ + AlCl₃

Part b.

According to the balanced equation shown above, the no. of moles of moles of silver chlorate produced from 1.00 moles of aluminum chlorate = 3.00 mol

Part a.

5 g AlCl₃ = 5 g/(133.5 g/mol) = 0.037453 mol

According to the balanced equation shown above, the mole ratio of AlCl₃:Al(ClO₃)₃ = 1:1

Therefore, the no. of moles of Al(ClO₃)₃ needed = 0.037453 mol

Now, 0.037453 mol Al(ClO₃)₃ = 0.037453 mol * 277.5 g/mol = 10.393 g