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?

Question

Na2S + Bi(NO3)3 -->Bi2S3 + Na^+(aq)+ NO3-(aq)( not balanced)

Answer 1

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.

Answer 2

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.