Answer to Question #312047 in Physical Chemistry for FSAFGDVCGTE3

Question

3 grams of Aluminum are reacted with 4 grams of Copper (II) sulfate to produce Aluminum sulfate

and the metal copper.

a) Which is the limiting agent?

b) How many grams of copper are formed

Answer 1

m(Al) = 3 g;

n(Al) = m(Al)/M(Al) = 3/27 = 0.11 moles;

m(CuSO₄) = 4 g;

n(CuSO₄) = m(CuSO₄)/M((CuSO₄) = 4/160 = 0.025 moles;

2Al + 3CuSO₄ = Al₂(SO₄)₃ + 3Cu;

n(Al)' = n(Al)/2 = 0.11/2 = 0.055 moles;

n(CuSO₄)' =n(CuSO₄)/3 = 0.025/3 = 0.008 moles;

So, aluminium (Al) is in excess and Copper (II) sulfate (CuSO₄) is the limiting agent.

For the chemical reaction:

n(Cu) = n(CuSO₄) = 0.025 moles;

m(Cu) = n(Cu) * M(Cu) = 0.025 * 64 = 1.6 g.

Answer: a) Copper (II) sulfate (CuSO₄) is the limiting agent;

b) 1.6 g.