How many grams of tin are required to form 1.0 g of silver by the following reaction? 2 Ag+(aq) + Sn(s) → 2 Ag(s) + Sn2+(aq)?

Method 1:

Molar mass of Sn = 118.7 g/mol
Molar mass of Ag = 107.9 g/mol

Balanced equation for the reaction:
2 Ag⁺(aq) + Sn(s) → 2 Ag(s) + Sn²⁺(aq)
Mole ratio Sn : Ag = 1 : 2

Number of moles of Ag formed = (1.0 g) / (107.9 g/mol) = 0.009268 mol
Number of moles of Sn required = (0.009268 mol) × (1/2) = 0.004634 mol
Mass of Sn required = (0.004634 mol) × (118.7 g/mol) = 0.550 g

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Method 2 :

(1.0 g Ag) × (1 mol Ag / 107.9 g Ag) × (1 mol Sn / 2 mol Ag) × (118.7 g Sn / 1 mol Sn)
= 0.550 g Sn (required)