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
see below.
i got the same answer
i think its right tho!
I got 9.76g. Isn't Bi(NO3)3 the limiting reactant? It has 0.03798mols and Na2S is in excess, 0.0584mols.
limiting reactant is Bi(NO3)3, and 9.762629805 g of Bi2S3 is produced
yeah the L.R is Bi(NO3)3 and the mass of Bi2S3 is 9.76g :)
To determine the maximum mass of bismuth sulphide (Bi2S3) that could precipitate and identify the limiting reactant, you need to use the stoichiometry of the balanced equation provided:
3Na2S + 2Bi(NO3)3 -> Bi2S3 + 6Na+ + 6NO3-
First, calculate the number of moles of each reactant:
Moles of Na2S = 4.55 g / molar mass of Na2S
Molar mass of Na2S = molar mass of Na + molar mass of S
Molar mass of Na = atomic mass of Na from periodic table
Molar mass of S = atomic mass of S from periodic table
Moles of Bi(NO3)3 = 15.0 g / molar mass of Bi(NO3)3
Molar mass of Bi(NO3)3 = (molar mass of Bi + 3 * (molar mass of N + (3 * molar mass of O))) from periodic table
Now, divide the moles of each reactant by their respective stoichiometric coefficients in the balanced equation to find the limiting reactant:
Moles of Na2S / 3 = 0.019333
Moles of Bi(NO3)3 / 2 = 0.019
From the calculations, it is evident that both moles are the same for Bi2S3 formation. Therefore, neither reactant is in excess, and there is no limiting reactant in this reaction.
Finally, calculate the mass of Bi2S3 that could form by multiplying the number of moles of Bi2S3 by its molar mass:
Mass of Bi2S3 = moles of Bi2S3 * molar mass of Bi2S3
Molar mass of Bi2S3 = (molar mass of Bi + molar mass of S) from periodic table
Substituting the values:
Mass of Bi2S3 = 0.019 * molar mass of Bi2S3
By performing the calculation, the maximum mass of bismuth sulphide (Bi2S3) is 9.77 g.
Therefore, the answer is 9.77 g, and as stated earlier, there is no limiting reactant in this reaction.