Question

Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation 3H2(g)+N2(g)→2NH3(g)

Part B

How many grams of NH3 can be produced from 2.15 mol of N2 and excess H2.

Part C

How many grams of H2 are needed to produce 13.02 g of NH3?

Part D

How many molecules (not moles) of NH3 are produced from 4.83×10⁻⁴ g of H2?

Express your answer numerically as the number of molecules.

Answer 1

B)

from reaction,

moles of NH3 = 2*moles of N2

= 2*2.15 mol

= 4.30 mol

Molar mass of NH3 = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

we have below equation to be used:

mass of NH3,

m = number of mol * molar mass

= 4.30 mol * 17.034 g/mol

= 73.2 g

Answer: 73.2 g

C)

Molar mass of NH3 = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass of NH3 = 13.02 g

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

= 13.02/17.034

= 0.7644 mol

From balanced chemical reaction, we see that

when 2 mol of NH3 produced, 3 mol of H2 is reacts

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

= (3/2)*0.7644

= 1.1465 mol

Molar mass of H2 = 2.016 g/mol

mass of H2 = number of mol * molar mass

= 1.1465*2.016

= 2.31 g

Answer: 2.31 g

D)

Molar mass of H2 = 2.016 g/mol

mass of H2 = 4.83*10^-4 g

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

= 4.83*10^-4/2.016

= 2.40*10^-4 mol

From balanced chemical reaction, we see that

when 3 mol of H2 reacts, 2 mol of NH3 is formed

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

= (2/3)*2.40*10^-4

= 1.60*10^-4 mol

Molecules of NH3 = moles of NH3*Avogadro’s number

= 1.60*10^-4 mol * 6.022*10^23 molecules/mol

= 9.64*10^19 molecules

Answer: 9.64*10^19 molecules