During studies of the reaction below,
2 n2h4(l) + n2o4(l) ? 3 n2(g) + 4 h2o(g)
a chemical engineer measured a less-than-expected yield of n2 and discovered that the following side reaction occurs.
n2h4(l) + 2 n2o4(l) ? 6 no(g) + 2 h2o(g)
in one experiment, 11.5 g of no formed when 102.1 g of each reactant was used.
what is the highest percent yield of n2 that can be expected?

The percent yield of the nitrogen gas is 11.53 %.

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

Given mass of NO = 11.5 g

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

Given mass of = 102.1 g

Molar mass of = 92 g/mol

Putting values in equation 1, we get:

For the given chemical reactions:

     ......(2)

      .......(3)

By Stoichiometry of the reaction 3:

6 moles of NO is produced from 2 moles of

So, 0.383 moles of NO will be produced from = of

By Stoichiometry of the reaction 2:

1 mole of produces 3 moles of nitrogen gas

So, 0.128 moles of will produce = of nitrogen gas

Now, calculating the experimental yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.384 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

By Stoichiometry of the reaction 2:

1 mole of produces 3 moles of nitrogen gas

So, 1.11 moles of will produce = of nitrogen gas

Now, calculating the theoretical yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 3.33 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

Experimental yield of nitrogen gas = 10.75 g

Theoretical yield of nitrogen gas = 93.24 g

Putting values in above equation, we get:

Hence, the percent yield of the nitrogen gas is 11.53 %.

can you list the answers so i can answer

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole