Question

A 500-g sample of Al2(SO4)3 is reacted with 450 g of Ca(OH)2. A total of 596 g of CaSO4 is produced. What is the limiting reagent in this reaction, and how many moles of excess reagent are unreacted?

Answer 1

k

Answer 2

give me a min

Answer 3

Al2(SO4)(aq) + 3Ca(OH)2(aq)---> 2Al(OH)3(s) + 3CaSO(s).

Answer 4

Al2(SO4)3 + 3Ca(OH)2 → 3CaSO4 + 2Al(OH)3
Al2(SO4)3 = 342.15g/mol
500g = 500/342.15 = 1.461 mol Al2(SO4)3
3*1.461 = 4.383 mol Ca(OH)2
Ca(OH)2 = 74.09 g/mol

Answer 5

It says AL2(SO4)3

AL - +3
SO4 - -2

Answer 6

sec

Answer 7

yes

Answer 8

ok

Answer 9

Ok

Answer 10

You want to see how much product each of the reactants will produce using the mole ratio from the balanced equation ..so....

Answer 11

yes

Answer 12

right?

Answer 13

i honestly dont know