Question

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

Answer 1

Answer: The enthalpy change of the reaction for given amount of manganese is 181.2 kJ

Explanation:

To calculate mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.00 g/mL

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of solution}}{100.0mL}\\\\\text{Mass of solution}=(1.00g/mL\times 100.0mL)=100g$

• To calculate the heat absorbed, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of solution = 100 g

c = specific heat capacity of solution = 4.18 J/g.°C

$\Delta T$ = change in temperature = $(28.2-23.0)^oC=5.2^oC$

Putting values in above equation, we get:

$q=100g\times 4.18J/g.^oC\times 5.2^oC\\\\q=2173.6J$

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of manganese = 0.650 g

Molar mass of manganese = 55 g/mol

Putting values in above equation, we get:

$\text{Moles of manganese}=\frac{0.650g}{55g/mol}=0.012mol$

• To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat absorbed = 2173.6 J = 2.174 kJ   (Conversion used:  1 kJ = 1000 J)

n = number of moles = 0.012 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{2.174kJ}{0.012mol}=181.2kJ/mol$

Hence, the enthalpy change of the reaction for given amount of manganese is 181.2 kJ