Limiting reactant = I₂
Mass of Mg leftover = 22.628 g
Further explanation
Given
Reaction between 25.0 g of Mg with 25.0 g of l₂
Required
The limiting reactant
mass of limiting and excess left over
Solution
Reaction
Mg + I₂⇒ MgI₂
mol Mg = 25 g : 24.305 g/mol = 1.029
mol I₂ = 25 g : 253.809 g/mol = 0.098
Limiting reactant = I₂
Excess reactant = Mg
The limiting reactant will be used up
Mol excess(unreacted) :
=1.029 - 0.098
=0.931
Mass I₂ left over :
= 0.931 x 24.305
= 22.628 g
