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!
