Question

Question 1 1 pts Suppose you are investigating the reaction: M(s) + 2 HCl(aq) → MCl2(aq) + H2(g). You weigh out a 0.241 gram piece of metal and combine it with 60.4 mL of 1.00 M HCl in a coffee- cup calorimeter. If the molar mass of the metal is 57 g/mol, and you measure that the reaction absorbed 131 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

Given that mass of the metal = .241g

First we have to find the limiting reactant.

Molar mass of the metal = 57g/mol

Therefore no.of moles of the metal present =0.241/57= 4.228*10^-3

Coming to Hcl, it is given that it is a 1.00 M sol. That means you will have 1 mole of HCl in 10³ ml of solution.

=>10^-3 moles of HCl for 1ml of solution.

So, for the given 60.4 ml of solution you will have 60.4*10^-3 moles of HCl.

According to the given reaction 1 mole of metal is required for reacting with 2 moles of HCl.

In the same way 30.2 * 10^-3 moles of metal are required for reacting with 60.4 * 10^-3 moles of HCl.

But moles of metal we have (4.228 * 10^-3 moles) is less than what is required. Therefore metal is the limiting reactant.

Next, for 4.228 * 10^-3 moles of limiting reactant (metal), heat absorbed was 131 J (given).

Now for one mole of metal the heat absorbed = 131/(4.228 * 10^-3)= 30.984 * 10³ J = 30.984 KJ

Therefore required enthalpy of reaction = -30.984KJ

It is negative since the reaction absorbs heat.