The combustion of 43.9g of ammonia produces g of no2

We calculate for the number of moles of ammonia using its molar mass:
     moles of NH3 = 43.9g NH3 (1mol NH3/17.031g NH3)
                             = 2.578 mol 
To compute for the number of moles of NO2, we use the mole ratio of ammonia and nitrogen dioxide from their coefficients in the balanced chemical equation of the combustion of NH3 in the presence of oxygen yielding nitrogen dioxide and water:
     4NH3(g) + 7O2(g) = 4NO2(g) + 6H2O(g)
which is 4 moles of ammonia is to produce 4 moles of nitrogen dioxide:
     moles NO2 = 2.578 mol NH3 (4mol NO2/4 mol NH3)
                        = 2.578 mol 
Using the molar mass of NO2, the mass of NO2 produced is therefore
     mass NO2 = 2.578 mol NO2(46.0055g NO2/1mol NO2)
                       = 118.6 grams NO2

Ok so co(g) is not pronoused that. the only clue is that youpronouse it the oppisote way. 

Using dilution equationm1v1 =m2v2 ; where m1 and m1 are the intiial and final concentration, and v1 and v2 are the initial and final volume.5  × 5 m = 8 l  × m2m2 = 25/8 = 3.125 mtherefore; the resulting molarity is 3.125 m