The chloride at 0, 272g of a sample was precipitated by the addition of 50.00 ml of AgNO3 0.1030 M. The titration
of excess silver, it has spent 8.65 ml of KSCN 0.1260 M. Express the outcome of this analysis Percentage of MgCl2
To calculate the percentage of MgCl2 in the given sample, we need to determine the amount of MgCl2 present in the sample and then calculate it as a percentage of the total sample mass.
Firstly, let's calculate the moles of AgNO3 used to precipitate chloride:
Moles of AgNO3 = molarity of AgNO3 × volume of AgNO3 solution
= 0.1030 M × 0.05000 L
= 0.00515 moles
AgNO3 reacts with chloride ions in a 1:2 ratio. So, moles of Cl- precipitated = 2 × moles of AgNO3 used
= 2 × 0.00515 moles
= 0.0103 moles
Next, let's calculate the moles of KSCN used to titrate the excess silver:
Moles of KSCN = molarity of KSCN × volume of KSCN solution
= 0.1260 M × 0.00865 L
= 0.0010899 moles
From the reaction, it is known that 1 mole of AgNO3 reacts with 1 mole of KSCN. Thus, the moles of AgNO3 remaining after titration = Moles of KSCN used
= 0.0010899 moles
To find the moles of AgNO3 reacted with MgCl2, subtract the remaining moles of AgNO3 from the total moles used:
Moles of AgNO3 reacted with MgCl2 = Total moles of AgNO3 used - Remaining moles of AgNO3
= 0.00515 moles - 0.0010899 moles
= 0.00406 moles
As per the reaction equation, 2 moles of AgNO3 react with 1 mole of MgCl2. So, the moles of MgCl2 present in the sample can be calculated as:
Moles of MgCl2 = (Moles of AgNO3 reacted with MgCl2) / 2
= 0.00406 moles / 2
= 0.00203 moles
Now, let's calculate the mass of MgCl2:
Mass of MgCl2 = Moles of MgCl2 × molar mass of MgCl2
= 0.00203 moles × (24.31 g/mol + 2 × 35.45 g/mol)
= 0.00203 moles × 95.21 g/mol
= 0.193 g
Finally, we can calculate the percentage of MgCl2 in the sample:
Percentage of MgCl2 = (Mass of MgCl2 / Mass of the sample) × 100
= (0.193 g / 0.272 g) × 100
= 71.03%
Therefore, the outcome of the analysis is that the sample contains approximately 71.03% MgCl2.