Aquantity of 2.00 × 102 ml of 0.461 m hcl is mixed with 2.00 × 102 ml of 0.231 m ba(oh)2 in a constant-pressure calorimeter of negligible heat capacity. the initial temperature of the hcl and ba(oh)2 solutions is the same at 21.66°c. for the process below, the heat of neutralization is −56.2 kj/mol. what is the final temperature of the mixed solutions? h+(aq) + oh−(aq) → h2o(l)

Answer : The final temperature of the mixed solutions will be,

Explanation :

First we have to calculate the moles of and .

Now we have to calculate the limiting reactant.

The balanced chemical reaction will be,

As, 1 mole of react with 2 mole of HCl

So, 0.0462 mole of react with mole of HCl

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of water.

From the balanced chemical reaction, we conclude that

As, 1 mole of react to give 2 moles of

So, 0.0462 mole of react to give moles of

The moles of water = 0.0924 mole

Now we have to calculate the heat released.

As, 1 mole of water releases heat = 56.2 KJ/mole

So, 0.0924 mole of water releases heat =

Now we have to calculate the final temperature of solution.

where,

q = amount of heat = 5192.88 J

= specific heat capacity =

The volume of water = 200 ml + 200 ml = 400 ml

As the density of water is, 1 g/ml.

So, the mass of water =

= final temperature = ?

= initial temperature =

Now put all the given values in the above formula, we get:

Therefore, the final temperature of the mixed solutions will be,