1.how many mols of SO2 could be produced from 16.0g of MgSO3 and 45.0ml of .125 HCL? 2.how many moles of excess reactant will remain?

1.
From Wikipedia : Molar mass of MgSO₃ = 104.3682 g/mol

MgSO₃ + 2HCl → MgCl₂ + SO₂ + H₂O

No. of moles of MgSO₃ = (16.0 g) / (104.3682 g/mol) = 0.1533 mol
No. of moles of HCl = (0.125 mol/L) × (45.0/1000) = 0.005625 mol
Obviously, the limiting reactant is HCl.

According to the equation, mole ratio HCl : SO₂ = 2 : 1
No. of moles of HCl reacted = 0.005625 mol
No. of moles of SO₂ produced = (0.005625 mol) × (1/2) = 0.00281 mol


2.
According to the equation, mole ratio MgSO₃ : HCl = 1 : 2
No. of moles of HCl reacted = 0.005625 mol
No. of moles of MgSO₃ reacted = (0.005625 mol) × (1/2) = 0.00281 mol
No. of moles of MgSO₃ remains = (0.1533 - 0.00281) mol = 0.150 mol