Question

1. How many grams of NH3 can be produced from 4.93 mol of N2 and excess H2.

2. How many grams of H2 are needed to produce 11.87 g of NH3?

3. How many molecules (not moles) of NH3 are produced from 7.69×10⁻⁴ g of H2?

Answer 1

m NH3 if 4.93 mol of N2 reacts:

N2 +3H2 --> 2NH3

1 mol of N2 will give you 2 mol of NH3

4.93 mol of N2 ---> 2*4.93 = 9.86 mol of NH3

MW = 17 g/mol

mass = mol*¨MW = 9.86*17 = 167.6 grams of NH3

2)

N2 +3H2 --> 2NH3

m = 11.87 g of NH3

mol = mass/MW = 11.87/17 = 0.6982 mol of NH3

2 mol o f NH3 ---> 3 mol of H2

ratio: 3/2

3/2*0.6982 =1.047 mol of H2

MW of H2 = 2

mass H2 = mol*MW = 1.047*2 = 2.09 grams of H2

3)

N2 +3H2 --> 2NH3

m = 7.69*10^-4 g of H2

mol H2 = mass/MW = (7.69*10^-4)/2= 3.85*10^-4

3 mol of H2 --> 2 mol of NH3

ratio

2/3

2/3*3.85*10^-4 = 2.566*10^-4 mol of NH3 will be produced

btu we need molecules so:

1 mol = 6.022*10^23 molecules

2.566*10^-4 mol =  2.566*10^-4 * 6.022*10^23 = 1.54*10^20 molecules of NH3