Question

how many moles of MgCl

ny moles of MgCl2 are present in 60.0 mL of a 0.123 M MgCl2 What is the molar concentration of the Mg2+ and Cr ions? 1) How many mo solution? What Solution?

Answer 1

(1) Number of moles of MgCl₂ is , n = Molarity x volume in L

= 0.123 M x 60.0 mL x10^-3 L/mL

= 7.38x10^-3 moles

MgCl₂ ----> Mg2+ + 2Cl-

1 mole of MgCl₂ produces 1 mole of Mg2+ & 2 moles of Cl-

So molar concentration of Mg2+ = [MgCl₂ ] = 0.123 M

Molar concentration of Cl- = 2x[MgCl₂ ] = 2x0.123 = 0.246 M

(2) The balanced reaction is NaOH + HCl ---> NaCl + H2O

Numb er of moles of NaOH , n = Molarity x volume in L

= 0.300 M x 10.0 mLx10^-3 L/mL

= 3x10^-3 moles

From the balanced equation ,

1 mole of NaOH reacts with 1 mole of HCl

3x10^-3 moles of NaOH reacts with 3x10^-3 moles of HCl

So volume of HCl , M = number of moles of HCl /molarity

= 3x10^-3 moles / 0.500 M

= 6x10^-3 L

= 6mL

The balanced reaction is Ba(OH)2 + 2HCl ---> BaCl2 + 2H2O

Numb er of moles of Ba(OH)2 , n = Molarity x volume in L

= 0.200 M x 10.0 mLx10^-3 L/mL

= 2x10^-3 moles

From the balanced equation ,

1 mole of Ba(OH)2 reacts with 2 mole of HCl

2x10^-3 moles of NaOH reacts with 2x2x10^-3 moles=4x10^-3 moles of HCl

So volume of HCl , M = number of moles of HCl /molarity

= 4x10^-3 moles / 0.500 M

= 4x10^-3 L

= 4mL