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

User Despina
by
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1 Answer

5 votes

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

Step-by-step explanation:

User Cheesemacfly
by
5.1k points