Answer: a) 
is the limiting reagent
b) 0.27 g of 
will be produced.
Explanation:
To calculate the moles :
According to stoichiometry :
3 moles of require = 2 moles of 
Thus 0.003 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
b) As 3 moles of give = 2 moles of
Thus 0.003 moles of give = of
Mass of 
Thus 0.27 g of will be produced.
Based on the given molecular formula, the molar mass of ethylene glycol is 62 g/mol. We solve for the number of moles of the solute,
n = (7.1 kg)(1000 g/ 1 kg) / 62g/mol = 114.52 mol
Then, we divide this value by the given mass of the water in kg
m = (114.52 mol) / 1.2 kg = 95.43 m
Thus, the molality of the substance is approximately equal to 95.43 m.
A “supersaturated” solution contains more dissolved material. supersaturated solutions lies in the temperature of the water. more sugar will dissolve in hot water than in cold. Meaning that by separating the 2, only the supersaturated sugar would dissolve leaving the regular sugar untouched.
Hope this helps and have a nice day!
It belongs to the “s-block”
