Question

AP Chemistry

Number of Moles Practice:

Part A) Lithium and nitrogen react to produce lithium nitride:6Li(s)+N2(g)→2Li3N(s). How many moles of lithium nitride are produced when 0.630 mol of lithium react in this fashion?   

Part B) How many moles of BCl3 are needed to produce 10.0 g of HCl(aq) in the following reaction?
BCl3(g) + 3 H2O(l) → 3 HCl(aq) + B(OH)3(aq)

Part C) How many moles of nitrogen are formed when 58.6 g of KNO3 decomposes according to the following reaction? The molar mass of KNO3 is 101.11 g/mol. 4 KNO3(s) → 2 K2O(s) + 2 N2(g) + 5 O2(g)

Answer 1

Part (A) :

Reaction:

6Li(s)+N2(g)→2Li3N(s).

Moles of Lithium = 0.630 mol Li.

Calculation of Li3N:

Moles of Li3N = Moles of Li x 2 mol Li3N / 6 Li

=0.630 mol Li x 2 mol Li3N/ 6 Li

= 0.21 mol Li3N

Part B:

BCl3(g) + 3 H2O(l) → 3 HCl(aq) + B(OH)3(aq)

Reaction: Mass of HCl = 10.0 g

Calculation of moles of HCl = 10.0 g / 36.46 g per mol

= 0.274 mol HCl

Moles of BCl3 = moles of HCl x 1 BCl3 / 3 mol HCl

= 0.274 mol HCl x 1 mol BCl3 / 3 mol HCl

=0.0914 mol BCl3

Part C :

4 KNO3(s) → 2 K2O(s) + 2 N2(g) + 5 O2(g)

Mass of KNO3 = 58.6 g

Moles of KNO3 = 58.6 g / 101.11 g per mol

= 0.5796 mol

Moles of N2 = moles of KNO3 x 2 mol N2 / 4 KNO3

= 0.5796 mol KNO3 x 2 mol N2 / 4 mol KNO3

=0.2898 mol