When 25.0 g of Mg with 25.0 g of l2, what was the limiting reactant? How many grams of the limiting reactant are left over? How many grams of the other reactant (in excess) is left over?

Limiting reactant = I₂

Mass of Mg leftover = 22.628 g

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

a:

by the combination of an unlimited number of elements