When 4.76 g of a solid mixture composed of NH4Cl and CaCl2 was dissolved in 100.0 mL of water, the temperature of the water rose by 4.73°C. How many grams of each substance was in the mixture?

Answer 1

This will take some work but I would look at setting up two simultaneous equations and solving them.
You know the heat is q = mass x specific heat x delta T ort
q = 100 x 4.184 x 4.73 = ? J. Change to kJ.
let X = grams NH4Cl
and Y = grams CaCl2. so
X + Y = 4.76 is the first equation.

Look up the heat of dissolution for CaCl2 and NH4Cl and set up the second equation from that in terms of X and Y. I think NH4Cl is about +15 kJ/mol(endothermic) and CaCl2 is about -80 kJ/mol but you need to look them up and use the correct value. .How many mols NH4Cl do you have? That's X/mm NH4Cl where mm is molar mass. mols CaCl2 is Y/mm CaCl2. So
heat solution NH4Cl*(X/mmNH4Cl) + heat sol CaCl2*(Y/mmCaCl2) = q in kJ/mol
Solve for X and Y in grams NH4Cl and CaCl2.
If you get stuck post ALL of your work.

Answer 2

Madapaka