Question

Suppose you are investigating the reaction: M(s) + 2 HCl(aq) → MCl2(aq) + H2(g). You weigh out a 0.295 gram piece of metal and combine it with 65 mL of 1.00 M HCl in a coffee-cup calorimeter. If the molar mass of the metal is 57.78 g/mol, and you measure that the reaction absorbed 104 J of heat, what is the enthalpy of this reaction in kJ per mole of limiting reactant? Enter your answer numerically to three significant figures in units of kJ/mol.

Answer 1

Answer: The enthalpy of the reaction is $5.30\times 10^{-4}kJ$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$   ....(1)

• For metal:

Given mass of metal = 0.295 g

Molar mass of metal = 57.78 g/mol

Putting values in above equation, we get:

$\text{Moles of metal}=\frac{0.295g}{57.78g/mol}=0.0051mol$

• To calculate the moles of a solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Volume of hydrochloric acid = 65 mL = 0.065 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 1 moles/ L

Putting values in above equation, we get:

$1mol/L=\frac{\text{Moles of hydrochloric acid}}{0.065L}\\\\\text{Moles of hydrochloric acid}=0.065mol$

• For the given chemical reaction:

$M(s)+2HCl(aq.)\rightarrow MCl_2(aq.)+H_2(g)$

By Stoichiometry of the reaction:

1 mole of metal reacts with 2 moles of hydrochloric acid..

So, 0.0051 moles of metal will react with = $\frac{2}{1}\times 0.0051=0.0102moles$ of hydrochloric acid.

As, given amount of hydrochloric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, metal is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

If 1 mole of metal absorbs 104 J of heat

Then, 0.0051 moles of metal will absorb = $\frac{104J}{1mol}\times 0.0051mol=0.530J$ of heat.

Converting joules to kilo joules, we use the conversion factor:

$1kJ=1000J$

Converting the given value in kilo joules, we get:

$\Rightarrow \frac{0.530J}{1000J}\times 1kJ=5.30\times 10^{-4}kJ$

Hence, the enthalpy of the reaction is $5.30\times 10^{-4}kJ$