Question

what is the maximum amount of nitrogen dioxide that can produce by mixing 4.2gof no and 3.2g of o2

Answer 1

Answer:

6.44g

Explanation:

2NO + O2 = 2NO2

Molecular weight of 2NO = 60g

Molecular weight of 2NO2 = 92g

Hence,

60g of NO produces 92g of NO2

So, 4.2g of NO produces NO2 = 4.2 x 92 / 60

= 6.44g

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Answer 2

Answer:

6.44g

Explanation:

Maximum amount of nitrogen dioxide to be produced to 6.44 gm.

First of all, we need to write the balance equation of the reaction

2NO + 02 --------------> 2NO2

Now, All we need to find out it's mole.

Given,

Weight of NO = 4.2 gm

So,

Mole = Weight/ M.mass Of NO

= 4.2/30

= 0.14  moles

In balance reaction, We take 2 mole of NO to yield 2 mole of NO2 .

So, Mole of No2 = 0.14 × 2/2

= 0.14  moles

Now,

Maximum amount of nitrogen dioxide (NO2) is nothing but find out its weight

Weight (NO2) = Mole of NO2 × M.mass of

NO2

= 0.14 × 46

= 6.44 gm