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:

Question

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

Answer 1

I worked this problem for someone last night. Convert 0.500 g KCl to grams AgCl, then subtract from total AgCl obtained (8.486 g). The difference is the AgCl due to MgCl2. Then convert that number (the difference) to grams MgCl2 and you have it.

Answer 2

To find the mass of the MgCl2 sample used, we need to use stoichiometry to relate the mass of AgCl produced to the moles of MgCl2 used.

First, let's calculate the number of moles of AgCl produced:
- The molar mass of AgCl is Ag = 107.87 g/mol + Cl = 35.45 g/mol = 143.32 g/mol
- The mass of AgCl produced is 8.486 g
- Dividing the mass by the molar mass, we get moles of AgCl:
Moles of AgCl = 8.486 g / 143.32 g/mol = 0.0592 mol

Next, let's use the balanced chemical equation to relate the moles of AgCl to the moles of MgCl2.
From the balanced equation, we can see that the ratio of AgCl to MgCl2 is 3:1.
- So, moles of MgCl2 = 0.0592 mol / 3 = 0.0197 mol

Now, let's calculate the mass of the MgCl2 sample:
- The molar mass of MgCl2 is Mg = 24.31 g/mol + 2(Cl) = 35.45 g/mol x 2 = 95.21 g/mol
- Using the moles of MgCl2 determined above, we can calculate the mass of the MgCl2:
Mass of MgCl2 = 0.0197 mol x 95.21 g/mol = 1.871 g

Therefore, the mass of the MgCl2 sample used to make the solid mixture is approximately 1.871 g.