Question

SrCl2*6H2O(s)---> SrCl2(s) + 6H2O(g)

2. At lower temperatures SrCl2*6H2O(s) dissociates partially. 2.800g of SrCl2*6H2O(s) is placed in a sealed 1.000-L flask with air at 25.0 C and 750.0 torr. When the flask is heated to 122.0 C some SrCl2*6H2O(s) dissoicates so that the total pressure is 1565.0 torr.


a. Calculate the pressure due to the H2O(g) produced by the dissociation reaction (Account for the dry air at higher temperature)


b. Calculate the moles of H2O produced by the reaction.



c. Calculate the percent of SrCl2*6H2O(s) in the flask that dissociates at this temperature

Answer 1

a) water that is evaporated creates additional pressure on the walls of the bulb
p = p2 – p1 = 208649,5Pa - 99991,8Pa = 108657,7Pa
b) p*V= n* R*T
n(H2O) = (p2*V)/(R*t2 ) = (208649,5Pa*1m3)/(8,31 J/molK*395K) = 63,57mol
c) SrCl2*6H2O(s)  SrCl2(s) + 6H2O(g)
nd(SrCl2*6H2O) = n(H2O)/6 = 63,57mol / 6 = 10,6mol
md (SrCl2*6H2O) = nd(SrCl2*6H2O)*M(SrCl2*6H2O) =
= 10,6mol *231 g/mol=2447g
η(SrCl2*6H2O) = (md(SrCl2*6H2O))/(m( SrCl2*6H2O)) * 100% = ( 2447g)/2800g *100% = 87%