Question

_{^{The first step in processing Zn metal from its ore, ZnS, is to react it with O2 according 2ZnS(s )+ 3O2(g)è2ZnO(s) + 2SO2(g)}}

_{^{If 875 kg of ZnS is to be reacted,}}

_{^{(a) How many moles of O2 must be present to completely consume all of the ZnS.}}

_{^{(b) What volume of O2 at 0.875 atm and 34.0 C is needed (at a minimum) to carry out this reaction?}}

Answer 1

                             2ZnS(s )              +    3O2(g)           --------------->        2ZnO(s) +          2SO2(g)

                    2*97.48 = 194.96g                    3*32 = 96g

                          875000g                               ?

O2 required for 875kg of ZnS to completely consume = 875000*96/194.96 = 430857.6118g

Converting it into moles, moles = weight / Molecular weight = 430857.6118/32 = 13464.3 moles

From ideal gas equation

PV = nRT

P = pressure = 0.875 atm

V = Volume = ? L

n = number of moles of gas = 13464.3mol

R = Gas constant = 0.0821 atm L mol^-1 K^-1

T = 273 + 34 = 307 K

V = nRT / P = 13464.3 *0.0821 * 307 / 0.875 = 387844.1625L or 387.844m³