Asked by Helly

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 )



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i m getting 9.77 g for Bi2S3 .
no limiting reactant .. both are same ..
can u plz conform
Answered by bobpursley
No limiting reactant? Odd.

What was the moles of Na2S?
What was the moles of Bi(NO3)3

Now balance the equation.
the coefficents tells you the mole ratio you need. What was the mole ratio you had?
Answered by bobsley plz conform
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
Answered by bobpursley
amazing. exactly the same. No limiting reactant.

check your calcs again. I might have made a mistake, but I got 9.57 g.

0.019*(2*208.98+3*32)
Answered by a
don't round the answers so early, and you will see that one of them is the limiting reactant
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