Based on the following balanced equation- how many grams of zinc oxide could be generated from 1.75 miles of zinc sulfide assuming there was excess oxygen?
c = 12.01, o = 16.00, zn = 65.38 (g/mol)

mass of ZnO = 142.42 g  

Explanation:

Balanced chemical equation  

       2ZnS + 3O₂ → 2ZnO  + 2SO₂

Given data from eq.

     moles of zinc sulfide = 2 mol  

     moles of zinc oxide = 2 mol  

     mass of zinc oxide = ?

     Given moles of  

zinc sulfide = 1.75 mol

Solution  

1st we will find out the mole ratio of zinc sulfide and zinc oxide from balanced chemical equation  

          ZnS : ZnO

               2 : 2

            2/2 : 2/2

                 1 : 1  

given number of moles of zinc sulfide = 1.75 mol

So we can find out moles (x) of ZnO  needed for 1.75 mol of zinc sulfide  

from the balanced chemical equation the mole ratios are:  

               1 : 1  

          1.75 : x  

Cross multiply these ratios  

        1 × 1.75 = 1x  

             1.75 = 1 x  

             x = 1.75 mol  

Now we will find out the mass of ZnO

    mass = moles × molar mass  

    mass of ZnO = 1.75 mol × 81.38 g/mol  

    mass of ZnO = 142.42 g  

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

the answer would be a