How many grams of aluminum will be produced if 21.3g of ba react with 19.7g of al2 (so4)3

2.78 g of Al will be produced

Explanation:

Let's think the reaction:

Al₂(SO₄)₃ + 3Ba  →  3BaSO₄  + 2Al

Now we should determine the limiting reactant.

We convert the mass to moles (Mass / Molar mass)

21.3g / 137.33 g/m = 0.155 moles of Ba.

19.7 g / 342.14 g/m = 0.0575 moles of Sulfate.

Ratio is 1:3, so 1 mol of sulfate reacts with 3 moles of Ba.

Then, 0.0575 moles of sulfate, will react with the x3 moles. (0.172 moles)

We do not have enough barium, so it is our limting reactant, we need 0.172 moles, but we only have 0.155 moles.

Now we can work with the reaction:

3 moles of Ba, produce 2 moles of Al

Then, 0.155 moles of Ba, will produce (0.155 .2)/ 3 = 0.103 moles.

Let's convert moles to mass ( mol . molar mass)

0.103 m . 26.98 g/m = 2.78 g

600 mL

Explanation:

molarity 1 * volume 1 = molarity 2 * volume 2. So this means 12 * 250 = 3000 and divide that by 5 to get 600 mL.