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

Molar weight of NaF is 42 g

Molar weight of Th(NO₃)₄ is 480.1 g

the balanced equation is

4NaF + Th(NO₃)₄ = 4NaNO₃ + ThF₄

Hence the ratio of NaF: Th(NO₃)₄ is 4:1 = 1:0.25

i.e. 1 mole of NaF will react with 0.25 mole of Th(NO₃)₄

1.26 g of NaF = 0.03 moles

7.8 g of Th(NO₃)₄ = 0. 016 moles

Hence, the given mole ratio of NaF: Th(NO₃)₄ = 0.03:0.016 = 1:0.533

0.03 moles of NaF wil react with  0.0075 moles (0.03 x 0.25 moles) of Th(NO₃)₄.

i.e. 0.016-0.0075 =  0.0085 moles of Th(NO₃)₄ is excess

So, the excess reagent is Th(NO₃)₄. To complete the reaction 0.0075 moles of Th(NO₃)₄ required. i.e. 3.6 g (0.0075 x 480.1 g) of Th(NO₃)₄ will react with 1.26 g of NaF.

The remaining amount of Th(NO₃)₄ = 7.6-3.6 = 4.8 g