Answer: The number of moles of excess reactant left over will be, 0.089 moles.
Explanation : Given,
Mass of = 20.5 g
Mass of = 13.8 g
Molar mass of = 30 g/mol
Molar mass of = 32 g/mol
First we have to calculate the moles of and
.
and,
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of
So, 0.683 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of excess reactant left over.
Number of moles of excess reactant left over = Given moles - Required moles
Number of moles of excess reactant left over = 0.431 mol - 0.342 mol
Number of moles of excess reactant left over = 0.089 mol
Therefore, the number of moles of excess reactant left over will be, 0.089 moles.