1.34g of magnesium are added to 120cm^3 of 0.200mol/dm^3 solution of silver nitrate..what mass of silver will be formed?

Molar mass of Mg = 24.31 g/mol
Molar mass of Ag = 107.9 g/mol

Mg(s) + 2AgNO₃ → Mg(NO₃)₂(aq) + 2Ag(s)

Initial number of moles of Mg = (1.34 g) / (24.31 g/mol) = 0.0551 mol
Initial number of moles of AgNO₃ = (0.200 mol/dm³) × (120/1000 dm³) = 0.0240 mol

According to the equation, mole ratio Mg : AgNO₃ = 1 : 2
If AgNO₃ completely reacts, no. of moles of Mg needed = (0.0240 mol) × (1/2) = 0.0120 mol < 0.0551 mol
Hence, Mg is in excess, and AgNO₃ is the limiting reactant (limiting reagent), i.e. completely reacts.

According to the equation, mole ratio AgNO₃ : Ag = 1 : 1
No. of moles of AgNO₃ reacts = 0.0240 mol
No. of moles of Ag formed = 0.0240 mol
Mass of Ag formed = (107.9 g/mol) × (0.0240 mol) = 2.59 g