Question

NaF reacts with Th(NO3)4 according to the following unbalanced chemical equation:

NaF + Th(NO3)4 → NaNO3 + ThF4

Calculate the mass in grams of the excess reagent remaining after the complete reaction of 1.26 g of NaF with 7.80 g of Th(NO3)4.

Answer 1

The balanced reaction between NaF and Th(NO3)4 is given as:

4NaF + Th(NO3)4 → 4NaNO3 + ThF4

Mass of NaF = 1.26 g

Molar mass of naF = 41.98 g/mole

Moles of NaF = 1.26 / 41.98

= 0.03

Mass of Th(NO3)4 = 7.80 g

Molar mass of naF = 480.05 g/mole

Moles of NaF = 7.80 / 480.05

= 0.016

Now,

From the equation it is clear that 4 moles NaF reacts with 1 mole Th(NO3)4 .

So, 0.03 moles NaF will react with = 1/4 * 0.03

= 0.0075 moles of Th(NO3)4

But we have 0.016 moles of Th(NO3)4

Therefore, Th(NO3)4 is in excess.

moles of Th(NO3)4 in excess = 0.016 - 0.0075

= 0.0085 moles

Mass of  Th(NO3)4 in excess = 0.0085 * 480.05

= 4.08 g