If 10.0 g of aluminum sulfite react with 10.0 g of sodium hydroxide determine how many grams of aluminum hydroxide will be produced

Answer 1

Al₂(SO₃)₃+6NaOH --> 2Al(OH)₃+3Na₂SO₃

turn mass into moles with molar mass

10.0 g Al2(SO3)3 ( 1 mol / 294.15 g) = 0.03399626 mol Al2(SO3)3

10.0 g NaOH (1 mol NaOH / 40.0 g) = 0.25 mol NaOH

turn moles into mol Al(OH)3 with coefficients:

0.03399626 mol Al2(SO3)3 (2 mol Al(OH)3 / 1 mol Al2(SO3)3)=0.0679925 mol Al(OH)3

0.25 mol NaOH (2 mol Al(OH)3 / 6 mol NaOH) = 0.083333 mol Al(OH)3

smaller theoretical yield is 0.0679 mol product, so that's the real theoretical yield, and Al2(SO3)3 is limiting

convert moles to grams with molar mass

0.0679925 mol Al(OH)3 (748.003 g Al(OH)3 / 1 mol Al(OH)3)=5.30359 g Al(OH)3

sig figs:

=5.30 g Al(OH)3