Question

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 togehter, what is the maximum mass of bismuth sulphide an insoulble compound that could precipitate? Which reactant was the limitting reactant?

Na2S + Bi(NO3)3 --- Bi2S3 + Na^+ + NP3^- ( not balanced )
---------------------------------------balanced equatoin is

3Na2S + 2Bi(NO3)3 --- Bi2S3 + 6Na^+ + 6NO3^-

moles of Na2S = 0.058
moles of Bi(no3)3 = 0.038

Na2S = 0.058 / 3 = 0.019
Bi(no3)3 = 0.038 / 2 = 0.019

both are same for Bi2S3 from balanced equation

so the mass is 9.77 g

no limiting reactant ..

plz conform

thanks

Answer 1

see below.

Answer 2

i got the same answer

i think its right tho!

Answer 3

I got 9.76g. Isn't Bi(NO3)3 the limiting reactant? It has 0.03798mols and Na2S is in excess, 0.0584mols.

Answer 4

limiting reactant is Bi(NO3)3, and 9.762629805 g of Bi2S3 is produced

Answer 5

yeah the L.R is Bi(NO3)3 and the mass of Bi2S3 is 9.76g :)