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
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?
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
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)
don't round the answers so early, and you will see that one of them is the limiting reactant
To determine the maximum mass of bismuth sulfide (Bi2S3) that could precipitate and identify the limiting reactant, we need to perform stoichiometric calculations.
First, we need to balance the chemical equation:
3Na2S + 2Bi(NO3)3 → Bi2S3 + 6NaNO3
Now, let's determine the number of moles of each reactant:
Number of moles of Na2S = mass of Na2S / molar mass of Na2S
= 4.55 g / (22.99 g/mol + 32.06 g/mol)
≈ 0.0879 mol
Number of moles of Bi(NO3)3 = mass of Bi(NO3)3 / molar mass of Bi(NO3)3
= 15.0 g / (208.98 g/mol + 3 * 14.01 g/mol + 9 * 16.00 g/mol)
≈ 0.0486 mol
According to the balanced equation, the stoichiometric ratio between Na2S and Bi(NO3)3 is 3:2. Therefore, the limiting reactant is the one that has a lower number of moles. In this case, Bi(NO3)3 is the limiting reactant because it has fewer moles than Na2S.
To determine the maximum mass of Bi2S3 that could precipitate, we use the stoichiometric ratio between Bi(NO3)3 and Bi2S3, which is 2:1:
Number of moles of Bi2S3 = 0.0486 mol * (1 mol Bi2S3 / 2 mol Bi(NO3)3)
≈ 0.0243 mol
Mass of Bi2S3 = number of moles of Bi2S3 * molar mass of Bi2S3
= 0.0243 mol * (208.98 g/mol + 32.06 g/mol)
≈ 9.78 g
So, the maximum mass of bismuth sulfide (Bi2S3) that could precipitate is approximately 9.78 g. The limiting reactant is Bi(NO3)3.