Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%
Explanation:
To calculate the moles :


According to stoichiometry :
1 mole of
require 3 moles of 
Thus 2.8 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 = 2 moles of 
Thus 2.8 moles of
give =
of 
Mass of 
Theoretical yield of liquid iron = 313.6 g
Experimental yield = 288 g
Now we have to calculate the percent yield

Therefore, the percent yield is, 91.8%
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To answer the question above, let us a basis of the 1000 mL or 1 L.
volume = (0.9928 g/mL)(1000mL) = 992.8 g
Then, determine the mass of the alcohol by multiplying the total mass by the decimal equivalent of 5%.
mass of alcohol = 0.05(992.8 g) = 49.64 g
Then, determine the number of moles of ethyl alcohol by dividing the mass of alcohol by the molar mass (46 g/mol).
n = 49.64 g/ (46 g/mol) = 1.08 mol
Then, divide the number of moles by the volume (our basis is 1 L)
molarity = 1.08 mol/ 1 L = 1.08 M