Question

1. Ammonia can also be synthesized by this reaction: 3H2(g)+N2(g) → 2NH3(g) What maximum amount of ammonia in grams can be synthesized from 25.2 g of N2 and 8.42 g of H2? Express your answer in grams to one decimal place.

2. What maximum amount of ammonia in kilograms can be synthesized from 5.22 kg of H2and 31.5 kg of N2? Express your answer in kilograms to one decimal place.

3. For the reaction 2Li(s)+F2(g) → 2LiF(s) identify the limiting reactant for each of the initial quantities of reactants.

Drag the appropriate items to their respective bins.

Items:

2.85X10^3 gLi

6.79X10^3 gF2

10.5gLi

37.2gF2

1.0gLi

1.0gF2

Li is the limiting reagent	F2 is the limiting reagent

Answer 1

1)

Molar mass of N2 = 28.02 g/mol

mass(N2)= 25.2 g

use:

number of mol of N2,

n = mass of N2/molar mass of N2

=(25.2 g)/(28.02 g/mol)

= 0.8994 mol

Molar mass of H2 = 2.016 g/mol

mass(H2)= 8.42 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(8.42 g)/(2.016 g/mol)

= 4.177 mol

Balanced chemical equation is:

N2 + 3 H2 ---> 2 NH3

1 mol of N2 reacts with 3 mol of H2

for 0.8994 mol of N2, 2.698 mol of H2 is required

But we have 4.177 mol of H2

so, N2 is limiting reagent

we will use N2 in further calculation

Molar mass of NH3,

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

= 1*14.01 + 3*1.008

= 17.034 g/mol

According to balanced equation

mol of NH3 formed = (2/1)* moles of N2

= (2/1)*0.8994

= 1.799 mol

use:

mass of NH3 = number of mol * molar mass

= 1.799*17.03

= 30.64 g

Answer: 30.6 g

2)

Molar mass of N2 = 28.02 g/mol

mass(N2)= 31500.0 g

use:

number of mol of N2,

n = mass of N2/molar mass of N2

=(3.15*10^4 g)/(28.02 g/mol)

= 1.124*10^3 mol

Molar mass of H2 = 2.016 g/mol

mass(H2)= 5220.0 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(5.22*10^3 g)/(2.016 g/mol)

= 2.589*10^3 mol

Balanced chemical equation is:

N2 + 3 H2 ---> 2 NH3

1 mol of N2 reacts with 3 mol of H2

for 1.124*10^3 mol of N2, 3.373*10^3 mol of H2 is required

But we have 2.589*10^3 mol of H2

so, H2 is limiting reagent

we will use H2 in further calculation

Molar mass of NH3,

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

= 1*14.01 + 3*1.008

= 17.034 g/mol

According to balanced equation

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

= (2/3)*2.589*10^3

= 1.726*10^3 mol

use:

mass of NH3 = number of mol * molar mass

= 1.726*10^3*17.03

= 2.94*10^4 g

= 29.4 kg

Answer: 29.4 kg

Only 1 question at a time please