Question

A student enters the lab and conducts Part A of the Experiment. The student uses 25.00 mL of 2.112 M HCl, and adds NaOH in excess as instructed. If the ΔH of the neutralization reaction is known to be -57,320 J/mol H2O, what is the total theoretical heat released (in Joules)?

Answer 1

Answer: 3026.5 J

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in L)}}` .....(1)

Molarity of
`HCl` solution = 2.112 M

Volume of solution = 25.00 mL

Putting values in equation 1, we get:

a)
`2.112M=\frac{\text{Moles of}HCl* 1000}{25.00ml}\\\\\text{Moles of }HCl=(2.112mol/L* 25.00)/(1000)=0.05280mol`

The balanced chemical reaction is:

`HCl+NaOH\rightarrow NaCl+H_2O`

As
`HCl` is the limiting reagent , it limits the formation of product.
`NaOH` is the excess reagent.

1 mole of HCl produces = 1 mole of
`H_2O`

0.05280 mole of HCl produces =
`(1)/(1)* 0.05280=0.05280` moles of
`H_2O`

Given :

Energy released when 1 mole of
`H_2O` is produced = 57320 J

Thus Energy released when 0.05280 moles of
`H_2O` is produced =
`(57320J)/(1)* 0.05280=3026.5J`

Thus 3026.5 J is the total theoretical heat released