Question

just need question #6 please !!!!

5) Suppose you place 0.500 g of Mg metal in a coffee cup calorimeter.You add 100.0 mL of 1.00 M HCl.The mass of the HCl solution is 100.112 g.The reaction that occurs is as follows:

Mg (s) + 2 HCl (aq) ® MgCl₂ (aq) + H₂ (g)

The initial temperature of the HCl solution is 22.2°C and the final temperature after the reaction is complete is 44.8°C. The specific heat of water is 4.184 J/g·K, and magnesium is the limiting reactant.

a. How much heat is given off (generated) in this reaction

b. What is the enthalpy change (DH) for the reaction per mole of magnesium

6*****) A 2.358 g sample containing iron(II) is titrated using the standardized solution from question 5.If the titration requires 33.55 mL of titrant to reach the endpoint, what is the mass percent of iron in the sample?

Answer 1

Heat generated= mass of HCl* specific heat of solution* ( final temperature- initial temperature)

=100.112*4.184*(44.8-22.2) =9466 J

this much heat is generated when 0.5 gm of Mg metal is reacted. ( since Mg is limiting reactant, all the Mg is consumed).

moles of Mg= mass of Mg/Atomic weight of Mg= 0.5/24=0.0208

enthalpy change/mole of Mg= 9466/0.0208 J/mole=455096 J/mole =455.096 KJ/mole

3. The reaction is Fe+2HCl------->FeCl2+H2

1 mole of Fe requires 2 moles of HCl

moles of HCL in 33.55ml of 1M= Molarity* Volume of HCL in liters= 1*33.5/1000=0.0335 moles

Moles of Fe required as per this information =0.0335/2= 0.01675

mass of Fe = moles of Fe* atomic weight= 0.01675*56= 0.938 gm

mass % of iron=100*( mass of iron/total mass)=100*0.938/2.358=39.77%