If 0.813 g of O3 reacts with 0.605 g of NO, how many grams of NO2 will be produced?

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:

there all right good

explanation:

they it depends on if one believes about radioactivity, they make a geiger counter beep but maybe it was going to beep anyway electronics do wierd things all the time or it was programmed to beep near certain metals.