If 4.55 g of sodium sulphide and 15.0 g of bismuth nitrate are dissolved in separate beakers of water which are then poured together, what is the maximum mass of bismuth sulphide, an insoluble compound, that could precipitate? What reactant was the limiting reactant?
Na2S + Bi(NO3)3 -->Bi2S3 + Na^+(aq)+ NO3-(aq)( not balanced)
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DrBob222, Thursday, January 14, 2010 at 4:53pm
1. Balance the equation.
2. Convert 4.55 g Na2S and 15.0 g Bi(NO3)3 to moles. moles = grams/molar mass.
3a. Using the coefficients in the balanced equation, convert moles Na2S to moles or the product.
3b. Do the same for Bi(NO3)3.
3c. Now you have two answers for moles Bi2S3 produced. Obviously on of them is correct and the other is not correct. The correct value ALWAYS is the smaller value.
4. Use the value from 3c and convert moles Bi2S3 to grams. grams = moles x molar mass.
Post your work if you get stuck.
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Saira, Thursday, January 14, 2010 at 8:29p
Is this Correct?
3Na2S + 2Bi(NO3)3 -->Bi2S3 + 6Na^+(aq)+ 6NO3-(aq)
A.) 4.55g/ 78.05= 0.059 moles Na2S
15.0g/ 394.99= 0.038 moles Bi(NO3)3
B.) Na2S= 0.059/ 3= 0.019
Bi(NO3)3= 0.038/ 2= 0.019
(Just for general knowledge if this was a different question, and after dividing by the coefficient one number was smaller than the other, would i use the smaller number for the next step, to get the mass)
C. ) 0.019 moles Bi2S3
g= moles* molar mass
0.019 * 514.16
= 9.76 g
So the mass of Bi2S3 is 9.76g and the limiting reactant is Bi(NO3)3 because it was 0.038 moles.
Almost but not quite. This may appear to be picky but I am just trying to be accurate. My comments appear in bold.
A.) 4.55g/ 78.05= 0.059 moles Na2S
15.0g/ 394.99= 0.038 moles Bi(NO3)3
Actually, this is 4.55/78.05 = 0.0583. You are allowed 3 places (by the 4.55) so you should use all three places. I used 78.046 but the answer is the same to three places.
15.0/394.99 = 0.03798 which COULD be rounded here to 0.0380 but I prefer to leave all of that in my calculator and round in the last step. If you round as each step is complete your work will sometimes contain rounding errors.
B.) Na2S= 0.059/ 3= 0.019
Bi(NO3)3= 0.038/ 2= 0.019
I will finish below AND answer your question in parentheses at the same time.
(Just for general knowledge if this was a different question, and after dividing by the coefficient one number was smaller than the other, would i use the smaller number for the next step, to get the mass)
C. ) 0.019 moles Bi2S3
g= moles* molar mass
0.019 * 514.16
= 9.76 g
So the mass of Bi2S3 is 9.76g and the limiting reactant is Bi(NO3)3 because it was 0.038 moles.
0.0583 x (1/3) = 0.0194 moles Bi2S3.
0.03798 x (1/2) = 0.01899 moles Bi2S3.
Therefore, to 3 s.f., 0.0194 rounds to 0.0194 for the Na2S and 0.01899 rounds to 0.0190 for Bi2S3.
Thus, the SMALLER number is 0.0190 moles Bi2S3 and that x 514.16 = 9.769 g Bi2S3, which rounds to 9.77 g Bi2S3 to 3 s.f. That is VERY close to your answer but you rounded the 0.0583 moles first calculated to 0.059 (actually, you rounded incorrectly since 0.0583 should round to 0.058 to 2 s.f.) and that makes a small difference. You are correct that Bi(NO3)3 is the limiting reagent but not for the reason you gave. It is the limiting reagent because it produces the smaller amount of Bi2S3; i.e., all of the Bi(NO3)3 will be used before the Na2S is completely used.
Yes, your calculations are correct.
To determine the maximum mass of bismuth sulphide that could precipitate, you correctly balanced the equation:
3Na2S + 2Bi(NO3)3 → Bi2S3 + 6Na+ + 6NO3-
Then, you calculated the moles of sodium sulphide (Na2S) and bismuth nitrate (Bi(NO3)3):
Na2S: 4.55 g / 78.05 g/mol = 0.0585 moles
Bi(NO3)3: 15.0 g / 394.99 g/mol = 0.0379 moles
Next, you divided the moles of each reactant by their respective coefficients in the balanced equation:
Na2S: 0.0585 / 3 = 0.0195 moles
Bi(NO3)3: 0.0379 / 2 = 0.0190 moles
It's worth noting that if one of the values obtained after this division was smaller than the other, you would use the smaller value in the next step to determine the mass.
Then, you calculated the moles of Bi2S3 produced using the correct value from the previous step (which is the smaller value):
0.0190 moles of Bi2S3
Finally, you determined the mass of Bi2S3 by multiplying the moles by the molar mass of Bi2S3 (514.16 g/mol):
0.0190 moles x 514.16 g/mol = 9.76 g
Therefore, the maximum mass of bismuth sulphide that could precipitate is 9.76 g, and the limiting reactant is Bi(NO3)3 because it was the reactant that resulted in the smaller number of moles of Bi2S3.