If a reaction vessel initially contains 5 mol S and 10 mil O2, how many moles of S will be in the reaction vessel once the reactants have reacted as much as possible

Explanation:

2S       +       3O₂       =       2SO₃

2moles      3 moles

2 moles of S react with 3 moles of O₂

5 moles of S will react with 3 x 5 / 2 moles of O₂

= 7.5 moles of O₂ .

O₂ remaining unreacted = 10 - 7.5 = 2.5 moles .

All the moles of S will exhausted in the reaction and 2.5 moles of oxygen will be left .  

its mg hope this

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