One step in the purification of zinc from ZnS ore is the following reaction with

Question

One step in the purification of zinc from ZnS ore is the following reaction with oxygen.  

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

A 20.0-L flask initially contains O2 at a pressure of 694 torr and a temperature of 35.0°C. Then 20.18 g of solid ZnS is added, and the reaction above goes to completion. If the temperature at the end of the reaction is 57.0°C, what will be the final total pressure in the flask? (Assume that the volume(s) of any solids present are small enough to neglect, so that the gas volume is 20.0 L throughout the reaction.)

Explanation / Answer

Initially, find the number of moles of O2
using:
P*V=n*R*T
(694/760)*20 = n*0.0821*(273+35)
n=0.722 mol
Moles of O2 present = 0.722 mol

mass of ZnS added= 20.18 g
molar mass of ZnS = 97.5 g/mol
Number of moles of ZnS = mass/molar mass
                                                  = 20.18/97.5
                                                  =0.207 mol

2 mol of Zns react with 3 mol of O2
ZnS is limiting reagent
So amount of O2 that reacted = (3/2)*0.207 = 0.311 mol

moles of SO2 gas formed = moles of ZnS = 0.207 mol
moles of O2 unreacted = 0.722 mol - 0.311 mol = 0.411 mol

Finally,
Total number of moles of gas present = 0.411 mol + 0.207 mol = 0.618 mol
use:
P*V= n*R*T
P*20 = 0.618*0.0821*(273+57)
P=0.837 atm
   = 0.837*760 torr
   =636 torr
Answer: 636 torr

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