The reaction between nitric oxide (no) and oxygen to form nitrogen dioxide (no2) is a key step in photochemical smog formation:
2no(g) + o2(g) → 2no2(g)
how many grams of o2 are needed to produce 1.21 g of no2?

0.4209 g

Explanation:

Given Data:

Balanced given reaction is

2NO(g) + O2(g) → 2NO2(g)

Grams of oxygen (O2 ) = to find that produce 1.21 g of NO2

Solution:

Data showed by a balance equation

2NO(g)     +     O2(g) → 2NO2(g)

1 mole           1 mole                                    2 mole    

From the reaction it is clear that 1mole of oxygen give 2mole of NO2

From formula we can calculate weight in grams for oxygen as well as for NO2

Mole= weight in grams/molecular weight (mass)………..(1)

Calculation for the weight in gram for oxygen

Mole= weight in grams of O2 /molecular weight (mass) of O2…………. (2)

Put values in formula 2

1 Mole of oxygen = weight in grams of O2 / 16x2

1 Mole oxygen = weight in grams of O2 / 32

= Weight in grams of O2 = 32grams

Calculation for the weight in gram for NO2

Mole= weight in grams of NO2 /molecular weight (mass) of NO2…………. (2)

Put values in formula 2

2 Mole NO2 = weight in grams of NO2 / 14+ (16x2)

2 Mole NO2 = weight in grams of NO2 / 46

= Weight in grams of NO2 = 46x2

= Weight in grams of NO2 = 92grams

Now coming back to balance Reaction by representing moles in grams

               2NO(g)     +     O2(g) → 2NO2(g)

              1 mole            1 mole                                    2 mole    

                                      32g                                         92g

By applying unity formula

If 32g of O2 gives ………………….92g of NO2

So how many grams of oxygen will produce 1.21g of NO2

X gram of O2 = 32x 1.21/92

= X gram of O2 = 32x 1.21/92

= X gram of O2 = 0.4209 g

So 0.4209 g required to produce 1.21g of NO2