Answer: 0.147 grams
Step-by-step explanation:
To calculate the moles :


According to stoichiometry :
2 moles of
require = 2 moles of
will require=
of

Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 2 moles of
give = 3 moles of

Thus 0.049 moles of
give =
of

Mass of

Thus 0.147 g of
will be produced from the given masses of both reactants.