4.44 grams of silver are produced when 0.50 g of magnesium reacts with silver nitrate.
How to determine mass?
To determine the mass of silver produced when 0.50 g of magnesium reacts with silver nitrate, we can use the balanced chemical equation:
Mg + 2AgNO₃ → Mg(NO₃)₂ + 2Ag
From the balanced equation, 1 mole of magnesium reacts with 2 moles of silver nitrate to produce 2 moles of silver.
First, calculate the moles of magnesium used:
n(Mg) = mass/molar mass
= 0.50 g / 24.31 g/mol
= 0.0206 mol Mg
Next, use the mole ratio from the balanced equation to determine the moles of silver produced:
n(Ag) = 2 moles Ag / 1 mole Mg × 0.0206 mol Mg
= 0.0412 mol Ag
Finally, calculate the mass of silver produced:
mass(Ag) = n(Ag) × molar mass
= 0.0412 mol Ag × 107.87 g/mol
= 4.44 g Ag
Therefore, 4.44 grams of silver are produced when 0.50 g of magnesium reacts with silver nitrate.