Question

How many grams of NO can be produced if 205g of NO2 is mixed with 59.0g of H2O?

Answer 1

Molar mass of NO2,

MM = 1*MM(N) + 2*MM(O)

= 1*14.01 + 2*16.0

= 46.01 g/mol

mass(NO2)= 205.0 g

use:

number of mol of NO2,

n = mass of NO2/molar mass of NO2

=(2.05*10^2 g)/(46.01 g/mol)

= 4.456 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 59.0 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(59 g)/(18.02 g/mol)

= 3.275 mol

Balanced chemical equation is:

3 NO2 + H2O ---> NO + 2 HNO3

3 mol of NO2 reacts with 1 mol of H2O

for 4.456 mol of NO2, 1.485 mol of H2O is required

But we have 3.275 mol of H2O

so, NO2 is limiting reagent

we will use NO2 in further calculation

Molar mass of NO,

MM = 1*MM(N) + 1*MM(O)

= 1*14.01 + 1*16.0

= 30.01 g/mol

According to balanced equation

mol of NO formed = (1/3)* moles of NO2

= (1/3)*4.456

= 1.485 mol

use:

mass of NO = number of mol * molar mass

= 1.485*30.01

= 44.57 g

Answer: 44.6 g