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 )
according to the balanced equation, for each one mole of Na2S one mole ofBi(NO3)3 is needed.
So how many moles of each do you start with? The lower number will be the limiting reactant. THis number will also be the number of moles of Bi2S3 produced, convert that to grams.
i m getting 9.77 g for Bi2S3 .
no limiting reactant .. both are same ..
can u plz conform
To determine the maximum mass of bismuth sulphide that could precipitate and identify the limiting reactant, we need to follow these steps:
Step 1: Write the balanced chemical equation.
The balanced chemical equation for the reaction between sodium sulphide (Na2S) and bismuth nitrate (Bi(NO3)3) is:
3Na2S + 2Bi(NO3)3 → Bi2S3 + 6NaNO3
Step 2: Calculate the number of moles for each reactant.
Using the given mass of each reactant and their respective molar masses, we can calculate the number of moles.
- Sodium sulphide (Na2S):
Given mass = 4.55 g
Molar mass of Na2S = 2(22.99 g/mol) + 32.06 g/mol = 46.05 g/mol
Number of moles = Given mass / Molar mass = 4.55 g / 46.05 g/mol ≈ 0.099 mol
- Bismuth nitrate (Bi(NO3)3):
Given mass = 15.0 g
Molar mass of Bi(NO3)3 = 208.98 g/mol + 3(14.01 g/mol) + 9(16.00 g/mol) = 485.24 g/mol
Number of moles = Given mass / Molar mass = 15.0 g / 485.24 g/mol ≈ 0.031 mol
Step 3: Determine the limiting reactant.
To identify the limiting reactant, we compare the mole ratios of the reactants to their coefficients in the balanced equation. The reactant that has a smaller mole ratio compared to the coefficient is the limiting reactant.
From the balanced equation:
3 moles of Na2S react with 2 moles of Bi(NO3)3 to produce 1 mole of Bi2S3
Mole ratio of Na2S to Bi(NO3)3:
0.099 mol Na2S × (2 mol Bi(NO3)3 / 3 mol Na2S) ≈ 0.066 mol Bi(NO3)3
Since the mole ratio of Na2S to Bi(NO3)3 is greater than 1, we can conclude that Na2S is the excess reactant while Bi(NO3)3 is the limiting reactant.
Step 4: Calculate the maximum mass of bismuth sulphide produced.
Since the balanced equation tells us that 2 moles of Bi(NO3)3 produce 1 mole of Bi2S3, we can use the mole ratio to calculate the number of moles of Bi2S3 produced.
Number of moles of Bi2S3 = 0.031 mol (from Bi(NO3)3)
Now, using the molar mass of Bi2S3, we can calculate its mass.
Molar mass of Bi2S3 = 2(208.98 g/mol) + 3(32.06 g/mol) = 514.23 g/mol
Mass of Bi2S3 = Number of moles × Molar mass = 0.031 mol × 514.23 g/mol ≈ 15.96 g
So, the maximum mass of bismuth sulphide that could precipitate is approximately 15.96 grams. The limiting reactant, as previously determined, is bismuth nitrate (Bi(NO3)3).