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)
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.
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.
To find the maximum mass of bismuth sulfide (Bi2S3) that could precipitate, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction and determines the maximum amount of product that can be formed.
To begin, we need to determine the number of moles of each reactant.
1. Sodium sulphide (Na2S):
Given mass of Na2S = 4.55 g
Molar mass of Na2S = 46.04 g/mol
Number of moles of Na2S = mass / molar mass = 4.55 g / 46.04 g/mol = 0.0989 mol
2. Bismuth nitrate (Bi(NO3)3):
Given mass of Bi(NO3)3 = 15.0 g
Molar mass of Bi(NO3)3 = 485.09 g/mol
Number of moles of Bi(NO3)3 = mass / molar mass = 15.0 g / 485.09 g/mol = 0.0309 mol
Next, we need to balance the chemical equation:
2Na2S + 3Bi(NO3)3 → Bi2S3 + 6NaNO3
From the balanced equation, we can see that 2 moles of Na2S react with 3 moles of Bi(NO3)3 to form 1 mole of Bi2S3.
Now, let's compare the number of moles of each reactant to determine the limiting reactant:
- Na2S has 0.0989 moles.
- Bi(NO3)3 has 0.0309 moles.
Since the molar ratio between Na2S and Bi2S3 is 2:1, we can conclude that for every 2 moles of Na2S, we obtain 1 mole of Bi2S3.
Comparing the number of moles, we can see that the number of moles of Bi(NO3)3 is less than half of the number of moles of Na2S.
Therefore, the limiting reactant in this reaction is Bi(NO3)3.
To find the maximum mass of Bi2S3 that could precipitate, we need to calculate the mass of Bi2S3 formed from the limiting reactant.
Molar mass of Bi2S3 = 514.16 g/mol
Number of moles of Bi2S3 formed = (0.0309 mol Bi(NO3)3) / (3 mol Bi(NO3)3/1 mol Bi2S3) = 0.0103 mol Bi2S3
Mass of Bi2S3 formed = number of moles x molar mass = 0.0103 mol x 514.16 g/mol = 5.30 g
Therefore, the maximum mass of bismuth sulfide (Bi2S3) that could precipitate is 5.30 grams.