If 10.0 g of Al2(SO3)3 is reacted with 10.0 g of NaOH…

Answer 1

What's the actual question?

Here is the complete molecular equation:

Al₂(SO₃)₃ + 6NaOH ==> 2Al(OH)₃ + 3Na₂SO₃

Limiting reactant?

10.0 g Al₂(SO₃)₃ x 1 mol Al₂(SO₃)₃ / 150. g = 0.0667 moles (÷1 -> 0.0667)

10.0 g NaOH x 1 mol NaOH / 40 g = 0.25 moles (÷6 -> 0.0417) THUS, NaOH IS LIMITING

How much Al(OH)₃ is formed?

0.25 mol NaOH x 2 mol Al(OH)₃ / 6 mol NaOH = 0.083 mole Al(OH)₃ (to find mass multiply by molar mass)

How much Na₂SO₃ is formed?

0.25 mol NaOH x 3 mol Na₂SO₃ / 6 mol NaOH = 0.125 mol Na₂SO₃ (to find mass multiply by molar mass)