Respuesta :

Answer:

0.74  g O

3

​

=

48

0.74

​

=0.0154 mol O

3

​

0.67 g NO=

30

0.67

​

=0.0223 mol NO

O

3

​

 is the limiting reagent and NO is in excess =0.0223−0.0154=0.007 mol

Thus, O

3

​

 taken =NO

2

​

 formed =0.0154 mol NO

2

​

                             =0.0154×46 g NO

2

​

Explanation:

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