Question

Question 2 1 pts Suppose you are investigating the reaction: M(s) + 2 HCl(aq) → MCl2(aq) + H2(g). You weigh out a 0.202 gram piece of metal and combine it with 62.2 mL of 1.00 M HCl in a coffee-cup calorimeter. If the molar mass of the metal is 48.51 g/mol, and you measure that the reaction absorbed 111 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. Question 3 1 pts A 312 g sample of a metal is heated to 257.896 °C and plunged into 200 g of water at a temperature of 20.43 °C. The final temperature of the water is 79.548°C. Assuming water has a specific heat capacity of 4.184 J/g °C, what is the specific heat capacity of the metal sample, in J/g °C)? Assume no heat loss to the surroundings. Report your response to 3 digits after the decimal.

5.

Answer 1

According to HOMEWORKLIB RULES we have to answer only first one problem

The reaction is as follows:

M (s) +2 HCl ----- > MCl2 (aq) + H2(g)

Number of moles of M = amount in g / molar mass

= 0.202 g /48.51 g/ mole

= 0.00416 moles M

                 

Moles of HCl = molarity * volume in L

=1.00 * 62.2/1000

= 0.0622 Moles HCl

Required mole of HCl

0.00416 moles M * 2 mole HCl /1 mole M

= 0.00832 moles HCl

Remaiang mole of HCl = total moles – used moles

= 0.0622 -0.00832

= 0.05388 mole HCl

Thus the limiting agent is metal M

The limiting agent has following properties:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

Given that energy = 111 J

Enthalpy per moles = total energy / number of moles of limiting agent

=111J / 0.00416 moles M

=26682.7 J/ MOLE

= 26.7 KJ/ MOLE