Question

11.16 In each of the following chemical equations, determine the moles and mass (in g) of the reactant used when 3.00 moles of the substance highlighted in blue forms. (a) 3N2H4(1) 4NH3(g) + N2(8) (b) 4PC15(s) — P4(s) + 10C12(8) (c) 4PC13 (8) —P4(s) +6C12(8)

Answer 1

a)

From given balanced equation,

Mol of N2H4 required = (3/4)*mol of NH3 reacted

= (3/4)*3.00 mol

= 2.25 mol

Molar mass of N2H4,

MM = 2*MM(N) + 4*MM(H)

= 2*14.01 + 4*1.008

= 32.052 g/mol

use:

mass of N2H4,

m = number of mol * molar mass

= 2.25 mol * 32.05 g/mol

= 72.12 g

Answer: 72.1 g

B)

From given balanced equation,

Mol of PCl5 required = (4/10)*mol of Cl2 reacted

= (4/10)*3.00 mol

= 1.20 mol

Molar mass of PCl5,

MM = 1*MM(P) + 5*MM(Cl)

= 1*30.97 + 5*35.45

= 208.22 g/mol

use:

mass of PCl5,

m = number of mol * molar mass

= 1.2 mol * 2.082*10^2 g/mol

= 250 g

Answer: 250 g

C)

From given balanced equation,

Mol of PCl5 required = 4*mol of P4 reacted

= 4*3.00 mol

= 12.0 mol

Molar mass of PCl5,

MM = 1*MM(P) + 5*MM(Cl)

= 1*30.97 + 5*35.45

= 208.22 g/mol

use:

mass of PCl5,

m = number of mol * molar mass

= 12 mol * 2.082*10^2 g/mol

= 2.50*10^3 g

Answer: 2.50*10^3 g