When a sample of aqueous hydrochloric acid was neutralized with aqueous sodium hydroxide in a calorimeter, the temperature of 148.3 g of water surrounding the reaction increased by 5.43 degrees * C . If the specific heat of water is 1.00 cal/(g C) , calculate the quantity of energy in calories involved in this neutralization reaction. (Ignore the heat gain by the calorimeter)

805.3cal

Explanation:

All neutralization reactions produce heat when occurs.

The reaction of HCl with NaOH produce heat increasing the aqueous medium in 5.43°C. The equation of the released energy is:

Q = S×m×ΔT

Where Q is heat,

S is specific heat of the solution (1.00cal/g°C)

m is mass of water: 148.3g

ΔT is change in temperature: 5.43°C

Q = 1.00cal/g°C×148.3g×5.43°C

Q = 805.3cal

well in order to answer this question you have to provide more information < 3