0.5 g of ca is added to 50 ml of 1.0 m hcl. once reaction is complete, the temperature of the solution has risen from 25.2 c to 71.4 c.
a. calculate the heat gained by the solution.
b. heat lost by the reaction
c. calculate the heat of reacton.
i've gotten several answers and i think i am overthinking the question. i am having trouble finding the correct mass to use.

Let us assume that the solution has the same specific heat capacity as water, of 4.18 Joules per gram.

a) H = mCpΔT
Where m will be the mass of the solution. We do not incorporate the mass of Ca because it will dissolve and release some amount of Hydrogen gas; rendering its contribution negligible. Assuming solution density equal to water's 1 gram per cubic centimeter:
H = 50 x 4.18 x (71.4 - 25.2)
H = 9655.8 Joules are gained

b) Assuming no heat is lost to the surroundings,
Heat lost by reaction = Heat gained by soluton
Heat lost by reaction = 9655.8 Joules

c) Moles of Ca present = 0.5/40 = 0.0125
Heat of reaction is the energy released upon reaction of the quantities in the balanced equation of a reaction. In this case:
Ca + 2HCl → CaCl₂ + H₂
So 0.0125 moles produces 9655.8 Joules
1 mole will produce: 9655.8 / 0.0125
= 772.5 kJ

Ethene molecular fórmula= c2h4molecular weight = 29.046g/molhope it : )