A sample of solid KCl (potassium chloride) weighing 0.500 g is mixed with an unweighed sample of solid MgCl2 (magnesium chloride) and the mixture is then completely dissolved in water to form a clear solution. An aqueous solution of AgNO3 (silver nitrate) is then added to this mixture. A reaction occurs and insoluble solid AgCl (silver chloride) is precipitated. The equation for the reaction is as follows:
KCl(aq) + MgCl2(aq)+ 3AgNO3(aq) KNO3(aq)+ Mg(NO3)2(aq)+ 3AgCl(s)
When the reaction is over, both of the original chlorides, KCl and MgCl2, are completely used up. Analysis shows that 8.486 g of AgCl has been produced. Find the mass of the MgCl2 sample used to make the solid mixture.
Use the following atomic masses: Ag = 107.87, Cl = 35.45, K = 39.10, Mg = 24.31, N = 14.01
Wouldn't this work. Check my thinking.
Calculate how much AgCl can be pptd from 0.500 g KCl. That will be
0.500 g KCl x (molar mass AgCl/molar mass KCl) = ?? g AgCl.
Subtract 8.486 - ??g AgCl from the above calcn to arrive at the amount AgCl that must be due to MgCl2 alone.
Then convert g AgCl that must be due to MgCl2 to g MgCl2 and you have it solved. I did a very quick quickie on the numbers and came up with approximately 2.7 grams MgCl2 but you need to check that carefully.
I ended up getting 1.24 g?
My answer, after reviewing it, is 2.499 grams MgCl2.
I used 95.21 for molar mass MgCl2.
143.32 for molar mass AgCl
74.55 for molar mass KCl.
i keep getting 1.24 g i first coverted 0.500g KCl to g of AgCl g and i got 2.88 g. then i subtracted that from 8.486 and i got 5.602 AgCl. then i converted 5.602 g of AgCl to g of MgCl2 and i got 1.24 g.
are your numbers similar to that?
To find the mass of the MgCl2 sample used in the mixture, we can use the equation and the information given to calculate the moles of AgCl produced, and then use stoichiometry to relate it to the moles of MgCl2.
First, we need to calculate the moles of AgCl produced. We know the mass of AgCl is 8.486 g. We can use the molar mass of AgCl to find the moles:
Molar mass of AgCl = atomic mass of Ag + atomic mass of Cl
= 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol
Number of moles of AgCl = mass of AgCl / molar mass of AgCl
= 8.486 g / 143.32 g/mol
= 0.059 mol
Next, we need to use the stoichiometry of the balanced equation to relate the moles of AgCl to the moles of MgCl2. According to the balanced equation, the molar ratio between AgCl and MgCl2 is 3:1 (3 moles of AgCl react with 1 mole of MgCl2).
Molar ratio of AgCl to MgCl2 = 3 : 1
Number of moles of MgCl2 = Number of moles of AgCl / Molar ratio of AgCl to MgCl2
= 0.059 mol / (3/1)
= 0.059 mol / 3
= 0.020 mol
Finally, we can calculate the mass of the MgCl2 sample using its molar mass:
Molar mass of MgCl2 = atomic mass of Mg + 2 * atomic mass of Cl
= 24.31 g/mol + 2 * 35.45 g/mol
= 95.21 g/mol
Mass of MgCl2 = Number of moles of MgCl2 * Molar mass of MgCl2
= 0.020 mol * 95.21 g/mol
= 1.904 g
Therefore, the mass of the MgCl2 sample used to make the solid mixture is 1.904 g.