Answer: 16.32 g of
as excess reagent are left.
Step-by-step explanation:
To calculate the moles :
According to stoichiometry :
2 moles of
require = 1 mole of
Thus 0.34 moles of
will require=
of
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
Moles of
left = (0.68-0.17) mol = 0.51 mol
Mass of
Thus 16.32 g of
as excess reagent are left.