Question

125 grams of solid sodium hydroxide is combined with an excess of aqueous hydrochloric acid. The heat of neutralization for the formation of water -55.9 kJ/mole. The heat of solvation (it might be called the heat of solution) for sodium hydroxide is 41.0 kJ/mole What quantity of heat is released?

Answer 1

Answer:

303 kilo Joules is the quantity of heat is released.

Explanation:

Moles of NaOH = $\frac{125 g}{40 g/mol}=3.125 mol$

$NaOH+HCl\rightarrow NaCl+H_2O$

According to recation 1 mole of NaOh gives 1 Mole of water, then 3.125 moles of NaOH will give ;

$\frac{1}{1}\times 3.125 mol=3.125 mol$ of water

The heat of neutralization for the formation of water = -55.9 kJ/mol

Heat evolved on formation of 3.125 moles of water : Q

$Q= 3.125 mol\times -55.9 kJ/mol=-175 kJ$

The heat of solvation for sodium hydroxide = -41.0 kJ/mole

Heat evolved by solvation of 3.125 moles of NaOH = Q'

$Q'=3.125 mol\times -41.0 kJ=-128 kJ$

Total quantity of heat is released when NaOH and HCl are mixed:

Q + Q' = -175 kJ + (-128) kJ = -303 kJ

(negative sign indicates that energy released)

303 kilo Joules is the quantity of heat is released.