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Answer: The change in enthalpy for each mole of zinc involved in the reaction is 152.4 kJ/mol.

Explanation:

First, we need to determine the moles of Zn and HCl.

The molar mass of Zn is 65 g/mole

and,

Next, we must identify the limiting reagent and the excess reagent.

The chemical reaction given is:

According to the balanced reaction we find that

1 mole of reacts with 2 moles of

Thus, 0.0206 moles of react with moles of

This leads us to determine that is the excess reagent because the moles provided exceed the required moles, while is limiting and restricts product formation.

Now to find the enthalpy change for each mole of zinc reacting in this reaction.

From the reaction we gather that,[ [TAG_59]]

0.0206 moles of Zn yield heat = 3.14 kJ

This implies that 1 mole of Zn generates heat =

Hence, the enthalpy change per mole of zinc involved in this reaction amounts to 152.4 kJ/mol.

To determine the mass of the lead piece, we use the following equation: Q(heat) = mC delta T, where Q equals 78.0 j, M is the mass we want to find, C is the specific heat capacity (0.130 j/g/C), and delta T shows the temperature difference, set at 9.0 c. Rearranging the formula to solve for M gives us M = Q / c delta T. By substituting in the values, we conclude that M = 78.0 j / (0.130 j/g/C * 9.0 C), calculating this gives us a mass of 66.7 g of lead.

The balanced equation is:

NaCH₃COO + HCl → NaCl + HCH₃COO

Make