Question

A 50-L reaction vessel is charged with O2 and excess ammonia and the following reaction occurs: 4NH3 + 5O2 → 4NO + 6H2O If the average rate of consumption of O2 is 0.00032 mol/L*min, what mass of NO will be present after 15 minutes of reaction?

Answer 1

-d[O2] / dt = 0.00032 M / min

relation between O2 consumption and formation of NO

- 1/5 d[O2] / dt = + 1/4 d [NO] /dt

d [NO] /dt = - 4/5 d[O2] / dt

                   = 4/5 * 0.00032

                    = 2.56 x 10^-4 M / min

1 min ---------------------> 2.56 x 10^-4 M

5 min -------------------> 5 x 2.56 x 10^-4 M = 1.28 x 10^-3 M

molarity of NO = 1.28 x 10^-3 M

volume = 50 L

molarity x volume = moles

1.28 x 10^-3 x 50 = moles

moles = 0.064

molar mass of NO = 14 + 16 = 30 g/mol

mass = moles x molar mass

mass = 0.064 x 30

mass = 1.92 g

mass of NO = 1.92 g