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.

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