What is the limiting reactant if 8 g of Ba reacts with 2.8 g of Al2 (SO4) 3?

Al2 (SO4) 3 is the limiting reactant

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3Ba + Al2 (SO4) 3 → 2Al + 3BaSO4

Next, we shall determine the mass of Ba and the mass of Al2 (SO4) 3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Al2 (SO4) 3 = 2x27 + 3[32 + (16x4) ]

= 54 + 3[32 + 64]

= 54 + 3[96]

= 54 + 288 = 342g/mol

Mass of Al2 (SO4) 3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

411g of Ba reacted with 342g of Al2 (SO4) 3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above,

411g of Ba reacted with 342g of Al2 (SO4) 3.

Therefore, 8g of Ba will react with

= (8 x 342/411 = 6.66g of Al2 (SO4) 3.

From the calculations made above, we can see that it will take a higher mass of Al2 (SO4) 3 i. e 6.66g than what was given i. e 2.8g to react completely with 8g of Ba.

Therefore, Al2 (SO4) 3 is the limiting reactant and Ba is the excess reactant.