A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculated the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol. He next tried the same experiment with 100 ml of 1.0 M HCl and 100 ml of 1.0 M NaOH. The calculated molar enthalpy change of reaction for his second trial was: Group of answer choices

-54 kJ/mol

Explanation:

Given that:

A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculated the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol

i.e

50 ml of 1.0 M HCl +  50 ml of 1.0 M NaOH > -54 kJ/mol

If he repeat the same experiment with :

100 ml of 1.0 M HCl + 100 ml of 1.0 M NaOH. >

From The experiment; the molar enthalpy of change of the acid-base neutralization reaction will be -54 kJ/mol

This is because : The second reaction requires 50 ml in order to neutralize the reaction, then the remaining 50 ml will be excess, Hence, there is no change in the enthalpy of the reaction.

Similarly; we can assume that :

In the first reaction;  P moles of  is used to liberate Q kJ heat ; then  the change in molar enthalpy will be Q/P (kJ/mol).

SO; when he used 100 ml ;

then the amount of moles used is double, likewise the heat liberated will be doubled ;

So;

2P moles is used to liberate 2Q kJ heat ;

2P/2Q = Q/P ( kJ/mol) = -54 kJ/mol

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%