Question

30.0 mL of 0.20 M AgNO, are added to 100.0 mL of 0.10 M HCI in a thermally nsulated vessel. The following reaction takes place: Ag (aq)+ Cl (aq)AgCI (s) The two solutions were initially at 22.00°C and the final temperature was 22.80 C Calculate the heat of this reaction in k.Jimol of AgCI formed. Assume a combined mass of 120 g and a specific heat capacity of 4.18 JK-'g for the reaction mixture.

Answer 1

Answer : The heat of this reaction of AgCI formed will be, 66.88 KJ

Explanation :

First we have to calculate the heat of the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = amount of heat = ?

$c$ = specific heat capacity = $4.18J/g.K$

m = mass of substance = 120 g

$T_{final}$ = final temperature = $22^oC=273+22=295K$

$T_{initial}$ = initial temperature = $22.8^oC=273+22.8=295.8K$

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

$q=120g\times 4.18J/g.K\times (295.8-295)K$

$q=401.28J$

Now we have to calculate the number of moles of $AgNO_3$ and $HCl$.

$\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume of solution}$

$\text{Moles of }AgNO_3=0.20mole/L\times 0.03L=0.006mole$

$\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume of solution}$

$\text{Moles of }HCl=0.10mole/L\times 0.1L=0.01mole$

Now we have to calculate the limiting reactant.

The balanced chemical reaction will be,

$AgNO_3+HCl\rightarrow AgCl+HNO_3$

As, 1 mole of $AgNO_3$ react with 1 mole of HCl

So, 0.006 mole of $AgNO_3$ react with 0.006 mole of HCl

From this we conclude that, $HCl$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgCl.

The given balanced reaction is,

$Ag^++Cl^-\rightarrow AgCl$

From this we conclude that,

1 mole of $Ag^+$ react with 1 mole $Cl^-$ to produce 1 mole of $AgCl$

0.006 mole of $Ag^+$ react with 0.006 mole $Cl^-$ to produce 0.006 mole of $AgCl$

Now we have to calculate the heat of this reaction of AgCI formed.

As, 0.006 mole of AgCl produced the heat = 401.28 J

So, 1 mole of AgCl produced the heat = $\frac{401.28}{0.006}=66880J=66.88KJ$

Therefore, the heat of this reaction of AgCI formed will be, 66.88 KJ