A 25.0 mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL o

Question

A 25.0 mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M barium nitrate solution and this precipitation reaction occurs:
2 KCl(aq) + Ba(NO3)2(aq)-> BaCl2(s) + 2 KNO3(aq)
The solid BaCl2 is collected, dried, and found to have a mass of 2.45 g. Determine the limiting reactant, the theoretical yield, and the percent yield. A 25.0 mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M barium nitrate solution and this precipitation reaction occurs:
2 KCl(aq) + Ba(NO3)2(aq)-> BaCl2(s) + 2 KNO3(aq)
The solid BaCl2 is collected, dried, and found to have a mass of 2.45 g. Determine the limiting reactant, the theoretical yield, and the percent yield.
2 KCl(aq) + Ba(NO3)2(aq)-> BaCl2(s) + 2 KNO3(aq)
The solid BaCl2 is collected, dried, and found to have a mass of 2.45 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

Explanation / Answer

1. Balanced equation should be: 2 KCl(aq) + Ba(NO3)2(aq) ---> BaCl2(s) + 2KNO3(aq)

2. Balanced equation shows we need twice as moles of KCl as Ba(NO3)2 - a 2:1 ratio. Lets see how much we have of each. moles = Molarity x Vol
a.) KCl: 1.20-mol/L x 0.025-L = 0.03 mol
b.) Ba(NO3)2: .900-mol/L x .015-L = 0.0135 mol x 2 = 0.027
c.) since the KCl we need is 2 x mol of Ba(NO3)2, we see we have slightly more than 2 times so Ba(NO3)2 is the limiting reagent.

3. The mass of BaCl2 that should form is:
0.0135 mol Ba(NO3)2 x (1 mol BaCl2 / 1 mol Ba(NO3)2) x ( 208.3- g BaCl2 / 1 mol BaCl2) =
2.81-g BaCl2

4. % yield = 2.45-g / 2.81-g x 100% = 87.2% yield

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