Question

How much heat is produced when 100mL of 0.25 M HCl (Density 1.00g/mL) and 200 mL of 0.150 M NaOH (Densty 1.00g/mL) are mixed?

HCl + NaOH ? NaCl + H2O Ho298= -58kJ
If both solutions are the same temperatureand heat capaciy of the products is 4.19 j/gC, how much will the temperature increase? What assumption did you make in your calculation?

Answer 1

Step-by-step explanation:

The given data is as follows.

For HCl : Volume = 100 mL, Density = 1.00 g/ml, Molarity = 0.25 M

For NaOH : Volume = 200 mL, Density = 1.00 g/ml, Molarity = 0.150 M

Therefore, mass of HCl will be calculated as follows.

Density =
`(mass)/(volume)`

1.00 g/ml =
`(mass)/(100 mL)`

mass = 100 g

Also, number of moles =
`Molarity * Volume`

Hence, no. of moles of HCl =
`0.25 M * 0.001 L` (as 1 ml = 1000 L)

=
`0.25 * 10^(-3) mol`

On the other hand, mass of NaOH will be calculated as follows.

Density =
`(mass)/(volume)`

1.00 g/ml =
`(mass)/(200 mL)`

mass = 200 g

Also, number of moles =
`Molarity * Volume`

Hence, no. of moles of HCl =
`0.150 M * 0.002 L` (as 1 ml = 1000 L)

=
`0.3 * 10^(-3) mol`

Total mass = (100 g + 200 g) = 300 g

Since, HCl is the limiting reagent over here. So, heat produced by
`0.25 * 10^(-3) mol` will be calculated as follows.

Q =
`0.25 * 10^(-3) mol * \Delta H^(o)_(298)`

=
`0.25 * 10^(-3) mol * -58 kJ`

=
`14.5 * 10^(-3)` kJ

=
`14.5 * 10^(-3) * 10^(3)` J

Also, it is known that relation between Q and temperature change is:

Q =
`mC \Delta T`

Hence, putting the values into the above formula as follows.

Q =
`mC \Delta T`

14.5 J =
`300 g * 4.19 J/g ^(o)C * \Delta T`

`\Delta T` =
`0.0115 ^(o)C`

Thus, we can conclude that increase in temperature is
`0.0115 ^(o)C`.