Question

5) Adding 0.150 s of Mg metal to 95.00 ml of 1,000 M HCI in a coffiee-cup calorimeter leads to a temperature increase of 0.831 "C Assume that the specific heat and density of the resuhting solution water, 4.18 VE'C) and 1.00 g/mL, respectively and assume that no heat is lost to are equal to those o surroundings. What is the molar heat of reaction, AH to obtain partial credit.) for the reaction below? twrine equations in order a) (2 points) Is this process endothermic or esothermic (circle the correct answer)? b) (2 points) Calculate the amount of heat to)? of Nie by CI expressed n More

Answer 1

a)

Since temperature rises, the process is exothermic. In an exothermic reaction, heat is liberated.

b)

95.00 mL of HCl $\text { 95.00 mL } \times \text { 1.00 g/mL } = \text { 95.00 mL }$
$\text { 95.00 mL } \times \text { 1.00 g/mL } = \text { 95.00 g }$

Total mass of solution $= \text { 95.00 g } + \text { 0.150 g } = \text { 95.150 g }$

$\text { The amount of heat of reaction } = \text { mass } \times \text { temperature change } \times \text { specific heat }$

$\text { The amount of heat of reaction } = \text { 95.150 g } \times \text { 0.831 }^oC \times \text { 4.18 }J/(g^oC)$

$\text { The amount of heat of reaction } = \text { 330.5 J }$

c)

0.150 g Mg (Molar mass 24.3 g/mol) $\dfrac { \text { 0.150 g } }{ \text { 24.3 g/mol } } = \text { 0.006173 mol}$

95.00 mL of 1.000 M HCl $\dfrac { \text { 95.00 mL } }{ \text { 1000 mL/L } } \times \text { 1.000 M } = \text { 0.095 mol}$

0.006173 moles of Mg will react with $2 \times \text { 0.006173 mol}= \text { 0.0123 mol}$ of HCl but 0.095 moles of HCl are present. Hence, HCl is excess reagent and Mg is limiting reagent.

The heat of reaction per mole of Mg is

$\dfrac { \text { 330.5 J }}{ \text { 0.006173 mol} } = \text { 53543 J/mol}$

Convert the unit from J to kJ.

$\text { 53543 J/mol} \times \text { 0.001 kJ/J} = \text { 53.5 kJ/mol}$