Answer: a)
is the limiting reagent
b) 3.59 g
c) 11.6g
Step-by-step explanation:
To calculate the moles :


According to stoichiometry :
a) 9 moles of
require= 4 moles of
Thus 0.15 moles of
will require=
of

Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
b) As 9 moles of
give = 8 moles of

Thus 0.15 moles of
give =
of

Mass of

c) As 9 moles of
give = 3 moles of

Thus 0.15 moles of
give =
of

Mass of
