Asked by Ana
Regarding previous post: I don't understand what I'm doing. My book says the answer is .50 M Ba(NO3)2 but I'm not getting that.
Answers
Answered by
DrBob222
You didn't tell me your answer but I'll bet is is 1 M Ba(NO3)2. You're forgetting the dilution of each of the reagents by each other. You should have 1 mole Ba(NO3)2 left over and that in 2 L is 0.5 M. Is that what you're doing?
Answered by
bobpursley
Dr Bob is right: you have 1 mole Ba(NO3)2 in two liters of solution, that is by definitiona, .5M
Answered by
Ana
i made a mistake along the way but i cant find it to fix it. should i have done 4 mol/2 L instead of 2 mole/1 L
Answered by
DrBob222
No.
2AgNO3 + BaCl2 ==> 2AgCl + Ba(NO3)2
2 x 1 = 2 moles AgNO3
2 x 1 = 2 moles BaCl2
AgNO3 is limiting; that is
2 moles AgNO3 will use 1 mole BaCl2.
At equilibrium,
All of the AgNO3 has been used.
BaCl2 = 2 moles initially - 1 mole reacted = 1 mole remaining.
1 mole/2L = 0.5 M.
2AgNO3 + BaCl2 ==> 2AgCl + Ba(NO3)2
2 x 1 = 2 moles AgNO3
2 x 1 = 2 moles BaCl2
AgNO3 is limiting; that is
2 moles AgNO3 will use 1 mole BaCl2.
At equilibrium,
All of the AgNO3 has been used.
BaCl2 = 2 moles initially - 1 mole reacted = 1 mole remaining.
1 mole/2L = 0.5 M.
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