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Manganese reacts with hydrochloric acid to produce manganese(II) chloride and hydrogen gas. Mn(s)+2HCl(aq) ? MnCl2(aq)+H2(g)

Part A When 0.650g Mn is combined with enough hydrochloric acid to make 100.0 mL of solution in a coffee-cup calorimeter, all of the Mn reacts, raising the temperature of the solution from 23.0?C to 28.2?C. Find ?Hrxn for the reaction as written. (Assume that the specific heat capacity of the solution is 4.18 J/g?C and the density is 1.00 g/mL.) Express your answer using three significant figures. ?Hrxn =____ kJ

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Final answer:

The enthalpy change (?Hrxn) for the given reaction is calculated using the formula ?Hrxn = (q / n), where q is the heat absorbed or released by the reaction, and n is the number of moles of the limiting reactant. In this case, the heat absorbed or released (q) can be calculated using the formula q = m * C * ?T, where m is the mass of the solution, C is the specific heat capacity of the solution, and ?T is the change in temperature. Finally, ?Hrxn is determined by dividing the heat (q) by the number of moles of the limiting reactant.

Step-by-step explanation:

To calculate the enthalpy change (?Hrxn) for the given reaction, we can use the formula:

?Hrxn = (q / n)

where q is the heat absorbed or released by the reaction, and n is the number of moles of the limiting reactant. First, we need to calculate the heat absorbed or released (q) using the equation:

q = m * C * ?T

where m is the mass of the solution, C is the specific heat capacity of the solution, and ?T is the change in temperature. In this case, the mass of the solution (m) is equal to the volume of the solution (100.0 mL) multiplied by the density of the solution (1.00 g/mL). The change in temperature (?T) is equal to the final temperature minus the initial temperature. Substituting these values into the equation and solving for q gives us the heat absorbed or released by the reaction. Finally, we can calculate ?Hrxn by dividing the heat (q) by the number of moles of the limiting reactant, which can be obtained from the stoichiometry of the reaction.

Using the given values:

m = (100.0 mL) * (1.00 g/mL) = 100.0 g

?T = (28.2?C) - (23.0?C) = 5.2?C

q = (100.0 g) * (4.18 J/g?C) * (5.2?C) = 2181.6 J

n = (0.650 g Mn) / (54.938 g/mol) = 0.0118 mol

Therefore, ?Hrxn = (2181.6 J) / (0.0118 mol) = 185104 J/mol = 185.1 kJ/mol (rounded to three significant figures).

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