Question

3Ba+Al2(SO4)3 -->2Al+3BaSO4,

Answer 1

Answer:

13.5 * 10^-2 g

Explanation:

What we know:

Balanced Equation: 3Ba+Al2(SO4)3 -->2Al+3BaSO4,

Grams of Ba: 1

Grams of Al2(SO4)3: 1.8g

Calculate the # of moles of Ba and Al2(SO4)3:

1g Ba/137.3 = 7.3 *10^-3 mol Ba

1.8g Al2(SO4)3/ 342 = 5.3 *10^-3 mol Al2(SO4)3

Find the limiting reactant:

Ba has a coefficient of 3 in the balanced equation, so we divide the # of moles of Ba by 3 to get... 7.3 *10^-3 mol Ba/3 = 2.43 *10^-3

Al2(SO4)3 has a coefficient of 1, so if we divide by 1, we get the same number of 5.3 *10^-3

2.43 *10^-3 is smaller than 5.3 *10^-3, therefore Ba is the limiting reactant.

finally, we just find the number of moles of Al

The ratio of Al to Ba is 2:3 so...

7.3 * 10^-3 * (2/3) = 5 *10^-3 mol Al

CONVERT TO GRAMS

5 *10^-3 mol Al  * 27 = 13.5 * 10^-2 g

Hope that was helpful!