If 5.3g o2 reacts with nh3 to produce 3.1g no2, what would the percent yield of no2 be? . 4nh3(g) + 7o2(g) --> 4no2 + 6h2o(g)

Molar mass

O2 = 16 x 2 = 32.0 g/mol        NO2 = 14 + 16 x 2 = 46.0 g/mol

4 NH3 + 7 O2 = 4 NO2 + 6 H2O

7 x 32.0 g O2 > 4 x 46.0 g NO2
5.3 g O2 > ( mass NO2)

mass NO2 = 5.3 x 4 x 46.0 / 7 x 32.0

mass NO2 = 975.2 / 224

= 4.3535 g of NO2

4.3535 g NO2 100 %
3.1 g NO2 ( percent yield )

percent yield = 3.1 x 100 / 4.3535

percent yield = 310 / 4.3535

= 71.20%

hope this helps!

Given equation is : 4 NH3 + 7 O2 = 4 NO2 + 6 H2O
where 
5.3 g of 02 react with NH3 to produce 3.1 g NO2
To calculate % yield of NO2
First Find the Molar mass of O2 and NO2 as following

O2 = 16 x 2 = 32.0 g/mol        NO2 = 14 + 16 x 2 = 46.0 g/mol

from equation above , we know that 7 oxygen molecules react to produce 2 molecules of NO2

7 x 32.0 g O2 > 4 x 46.0 g NO2
5.3 g O2 +NH3 > ( mass NO2)

mass NO2 = 5.3 x 4 x 46.0 / 7 x 32.0

mass NO2 = 975.2 / 224

= 4.3535 g of NO2

4.3535 g NO2 100 %
3.1 g NO2 ( percent yield )
Percent yield of NO2 can be calcuated by :

percent yield of NO2 = 3.1 x 100 / 4.3535

percent yield = 310 / 4.3535

Percent yield of NO2= 71.20%