Ammonia can also be synthesized by this reaction: 3H2(g) + N2(g) 2NH3(g) What ma

Question

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?

What maximum amount of ammonia in kilograms can be synthesized from 5.22 kg of H2 and 31.5 kg of N2?

Express your answer in kilograms to one decimal place.

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

2Li(s) + F2(g) 2LiF(s)

1.0 g Li; 1.0 g F2

10.5 g Li; 37.2 g F2

2.85×103 g Li; 6.79×103 g F2

Explanation / Answer

A)

3H2 + N2 --------------- 2 NH3

mass of N2= 25.2 grams

molar mass of N2 = 28 gram/mole

number of moles of N2= 25.2/28= 0.9 mole

mass of H2= 8.42 grams

molar mass of H2 = 2 gram/mole

number of moles of H2= 8.42/2= 4.21 mole

according to euation

1mole of N2= 3 mole of N2

0.9 mole of N2 =?

                           = 3x0.9= 2.7 moles of H2

we need 2.7 mole of H2. but we have 4.21 moles of H2 . so H2 is excess reagent.

N2 is limiting reagent.

according to r=equation

1 mole of N2= 2 mole of NH3

0.9 mole of N2 = ?

                   = 2x0.9= 1.8 mole of NH3

moles of NH3 = 1.8 mole

molar mass of NH3 = 17 grams/mole

mass of 1.8 mole of NH3= 1.8x17= 30.6 grams of NH3

Mass of NH3 is formed= 30.6 grams

B)   N2 + 3 H2 -------- 2 NH3

mass of N2= 31.5 Kg = 31500 grams

number of moles of N2= 31500/28=1125 mole

mass of H2= 5.22 Kg= 5220 grams

number of moles of H2= 5220/2= 2610 moles

according equation 1mole of N2= 3 mole of H2

                             1125 mole of N2= ?

                                              = 1125 x3=3375 moles of H2

number of moles of H2= 3375 moles

we need 3375 mole of H2. but we have 2610 moles of H2. so it is used first. hence it is limiting reagent.

according to equation , 3 mole of H2= 2 mole of NH3

                               2610 mole of H2 = ?

                                                   = 2x2610/3=1740 moles of NH3

mass of 1740 mole of NH3 = 1740x17= 29580 grams of NH3

mass of NH3= 29580 grams= 29.580Kg= 29.6Kg

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