Answer: 7.29 g of
will be produced from the given masses of both reactants.
Step-by-step explanation:
To calculate the moles :


According to stoichiometry :
1 mole of
require = 8 moles of

Thus 0.0267 moles of
will require=
of

Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 1 mole of
give = 8 moles of

Thus 0.0267 moles of
give =
of

Mass of

Thus 7.29 g of
will be produced from the given masses of both reactants.