Question

Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation

3H2(g)+N2(g)→2NH3(g)

1a. How many grams of NH3 can be produced from 2.60 mol of N2 and excess H2

1b. How many grams of H2 are needed to produce 11.32 g of NH3?

1c. How many molecules (not moles) of NH3 are produced from 4.06×10⁻⁴ g of H2?

Answer 1

a)

From balanced reaction,

mol of NH3 formed = (2/3)*mol of N2 reacted

= (2/3)*2.60 mol

= 1.73 mol

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

use:

mass of NH3,

m = number of mol * molar mass

= 1.73 mol * 17.03 g/mol

= 29.47 g

Answer: 29.5 g

b)

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass of NH3 = 11.32 g

mol of NH3 = (mass)/(molar mass)

= 11.32/17.03

= 0.6646 mol

According to balanced equation

mol of H2 required = (3/2)* moles of NH3

= (3/2)*0.6646

= 0.9968 mol

Molar mass of H2 = 2.016 g/mol

mass of H2 = number of mol * molar mass

= 0.9968*2.016

= 2.01 g

Answer: 2.01 g

c)

Molar mass of H2 = 2.016 g/mol

mass of H2 = 4.06*10^-4 g

mol of H2 = (mass)/(molar mass)

= 4.06*10^-4/2.016

= 2.014*10^-4 mol

According to balanced equation

mol of NH3 formed = (2/3)* moles of H2

= (2/3)*2.014*10^-4

= 1.343*10^-4 mol

Number of NH3 molecules = mol of NH3 * Avogadro’s number

= 1.343*10^-4 * 6.022*10^23

= 8.09*10^19 molecules

Answer: 8.09*10^19 molecules