Question

A gas contains a mixture of NH3(g) and N2H4(g), both of which react with O2(g) to form NO2(g) and H2O(g). The gaseous mixture (with an initial mass of 61.00 g) is reacted with 10.00 moles O2, and after the reaction is complete, 4.062 moles O2 remains. Calculate themass percent of N2H4(g) in the original gaseous mixture.

Answer 1

Answer:

Mass percent of N2H4 in original gaseous mixture = 31.13 %

Explanation:

Given:

Initial mass of gaseous mixture = 61.00 g

Initial mole of oxygen = 10.0 mol

Moles of oxygen remaining after the reaction = 4.062 mol

Moles of oxygen used = 10.0 - 4.062 = 5.938 mol

$4NH_3 + 7O_2\rightarrow 4NO_2 + 6H_2O$

$N_2H_4 + 3O_2\rightarrow 2NO_2 + 2H_2O$

Total oxygen used in both the reactions = 10.0 parts

out of 10 parts, 3 part react with N2H4.

$No.\;of\;moles\;of \;oxygen \;used = 5.398\times\frac{3}{10} =1.78\; moles$

Now, consider the reaction of N2H4

$N_2H_4 + 3O_2\rightarrow 2NO_2 + 2H_2O$

3 moles of O2 react with 1 mole of N2H4

1.78 moles of oxygen will react with 1.78/3 = 0.5933 mol of N2H4

$Mass = Moles\times Molecular mass$

Molecular mass of N2H4 = 32 g/mol

$Mass\;of\;N_2H_4= 0.5933\times 32 = 18.99 g$

$Mass\;percent = \frac{Mass\;of\;N_2H_4}{Total\;mass}\times 100$

Total mass = 61.0 g

$Mass\;percent = \frac{18.99}{61.0}\times 100=31.13 \%$