Question

4 NO_2(g) + 6 H₂O_(g) --> 7 O_2(g) + 4 NH_3(g)

In the reaction above, how many grams of NH₃ will be made from 8.5 L of NO2 gas at a temperature of 355 C and 0.985 atm? Show your work.

In the reaction above, given the conditions described in Part E, how many moles of O₂ gas will be made? Show your work.

Answer 1

4 NO2 (g) + 6 H2O(g) --> 7 O2 (g) + 4 NH3 (g)

first calculate the moles of NO2 reacted.

PV = nRT

n = PV /RT = (0.985 x 8.5 / 0.0821 x 628) = 8.3725 / 51.5588

n = 0.162

0.162 moles of NO2 reacted.

now according to balanced reaction

4 x 46 g NO2 forms 4 x 17 g NH3

0.162 moles NO2 forms 0.162 x 4 x 17 / 4 x 46 = 11.016 / 184 = 0.0599 g

mass of NH3 formed = 0.0599 g

second part :

4 moles NO2 gives 7 moles O2

0.162 moles NO2 gives 0.162 x 7 / 4 = 0.567

moles of O2 formed = 0.567