The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below
write the reacting equation
Co(NO3)2 + Li2S = 2LiNO3 + COS
find the moles of CO(NO3)2 = molarity x volume
= 130 ml x 0.160=20.8 moles
since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles
volume of Lis is therefore = moles of Lis/ molarity of LiS
= 20.8/0.160 = 130 Ml
b,f,h are already balanced
The final temperature in Celsius of the metal block is 49°C.
<h3>How to find the number of moles ?</h3>
Moles water = 
= 
= 0.0266 moles
Heat lost by water = 0.0266 mol x 44.0 kJ/mol
= 1.17 kJ
= 1170 J [1 kJ = 1000 J]
Heat lost = Heat gained
Heat gained by aluminum = 1170 J
1170 = 55 x 0.903 (T - 25) = 49.7 T - 1242
1170 + 1242 = 49.7 T
T = 48.5°C (49°C at two significant figures)
Thus from the above conclusion we can say that The final temperature in Celsius of the metal block is 49°C.
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