Please help me ASAP. I\'ve been working on this for a few days now but I\'m not

Question

Please help me ASAP. I've been working on this for a few days now but I'm not understanding how to go about this. Please offer me guidance. Much appreciated.

M=Unknown Metal

Volume of M2CO3 = 50mL ; Final Volume of 0.10M HCl = 15.00 mL ; Initial Volume of 0.10M HCl = 2.81 mL ; Volume 0.10M HCl added = 12.19 mL

1. Using the data obtained, calculate the moles of th HCl used to neutralize the unknown Group 1 metal carbonate dissolvedin the 50 mL sample for each trial titration.

2. For each trial, calculate the total moles of the unknown Group 1 metal carbonate originally dissolved int he 500 mL of distilled or deionized water.

3. Calculate hte molar mass of the unknown Group 1 metal carbonate and identify the metal.

4. Calculate hte percent error in the experimental determination of the molar mass.

Explanation / Answer

HCl + M2CO3 -> MCl + CO2 +H2O

There is one to one ratio of carbonate to carbon dioxide,

n(CO2) = n(M2CO3)

For molar mass,

M(M2CO3) = m(M2CO3) / n(M2CO3)

= m(M2CO3) / n(CO2)

Unknown metal is a metal carbonate and the method is gas evolution by means of hydrochloric acid. The unknown metal is Alkali, there are 3 possibilities: lithium, sodium, and potassium as M

HCl + M2CO3 -> MCl + CO2 +H2O

Molar mass of the metal carbonate is given by,

M(M2CO3) = mass of metal carbonate / moles of carbon dioxide

= m(M2CO3) / n(CO2)

1.

nNaOH = MNaOH * VNaOH

= (0.05413 mol/L)*(0.02155L)

= 0.001166501 mol NaOH

0.001166501 mol NaOH * 1 mol HCl / 1 mol NaOH = 0.001166501 mol HCl

MHCL = nHCl / VHCl

= 0.001166501 mol HCl / 0.250 L HCl

= 0.00466006 mol/L

Therefore 0.001167 moles were origibally added , and after alkali reaction occured concentration of HCl was 0.00466006 mol/L

2.

nHCl = MHCl * VHCl

nHCl = ( 1.012 mol / L ) * ( 0.02000 L )

nHCl = 0.02024 moles of HCl added in Part I

nreacted = nadded - nexcess

nreacted = 0.02024 mol HCl - 0.001166501 mol HCl

nreacted = 0.019073499

Therefore 0.02024 moles of HCl were put into the volumetric flask, and 0.01907 moles reacted with the alkali carbonate.

3.

M2CO3 + 2 HCl -> 2 MCl + CO2 + H2O + unreacted HCl

   nM2CO3 = 0.02024 mol HCl * ( 1 mol M2CO3 / 2 mol HCl )
   nM2CO3 = 0.01012 mol M2CO3

   Therefore there were 0.01012 moles of the alkali carbonate added to the volumetric flask

4.

MM2CO3 = m M2CO3 / nM2CO3

   MM2CO3 = 0.4802 g / 0.01012 moles HCl

   MM2CO3 = 47.45059289 g / mol M2CO3

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.