Mass to mass: How many grams of barium chloride are needed to produce 89.0g of barium nitrate? AgNO3+BaCl2->AgCl+Ba(NO3)2 Thanks!?

Molar mass of BaCl₂ = (137.3 + 35.5×2) g/mol = 208.3 g/mol
Molar mass of Ba(NO₃)₂ = (137.3 + 14.0×2 + 16.0×6) g/mol = 261.3 g/mol

2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂
Mole ratio BaCl₂ : Ba(NO₃)₂ = 1 : 1

No. of moles of Ba(NO₃)₂ produced = (89.0 g) / (261.3 g/mol) = 0.3406 mol
No. of moles of BaCl₂ needed = 0.3406 mol
Mass of BaCl₂ needed = (0.3406 mol) × (208.3 g/mol) = 70.9 g

I got 70.6g BaCl. Thanks for your help and showing how you got your answers. This is greatly helping me understand these problems.