44.5 g Al2(SO4)3
Explanation:
# moles Ba(NO3)2 = (0.543M)(0.717 L)
= 0.389 mol Ba(NO3)2
Now we use the molar ratios,
0.389 mol Ba(NO3)2 ×(1 mol Al2(SO4)3/3 mol Ba(NO3)2)
= 0.130 mol Al2(SO4)3
0.130 mol Al2(SO4)3 × (342.15 g Al2(SO4)3/(1 mol Al2(SO4)3)
= 44.5 g Al2(SO4)3
