If 100.0 mL of 0.171 M Na2SO4 are added to 100.0 mL of 0.889 M Pb(NO3)2, how many grams of PbSO4 can be produced?
Equation: Na2SO4 + Pb(NO3)2 = 2NaNO3 + PbSO4
To determine the number of grams of PbSO4 that can be produced, we need to calculate the limiting reagent first. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
First, we need to determine the number of moles of each reactant:
For Na2SO4:
Molarity (M) = 0.171 M
Volume (V) = 100.0 mL = 0.1 L
Number of moles (n) = Molarity * Volume
= 0.171 M * 0.1 L
= 0.0171 moles
For Pb(NO3)2:
Molarity (M) = 0.889 M
Volume (V) = 100.0 mL = 0.1 L
Number of moles (n) = Molarity * Volume
= 0.889 M * 0.1 L
= 0.0889 moles
Next, we need to determine the stoichiometry of the reaction. From the balanced equation, we see that 1 mol of Na2SO4 reacts with 1 mol of Pb(NO3)2 to produce 1 mol of PbSO4.
Since the stoichiometry ratio is 1:1, the limiting reactant is the reactant that has the fewest moles. In this case, Na2SO4 has 0.0171 moles and Pb(NO3)2 has 0.0889 moles. Therefore, Na2SO4 is the limiting reactant.
Now, we can use the stoichiometry to calculate the moles of PbSO4 that can be produced from the limiting reactant:
From the balanced equation, we know that for every 1 mole of Na2SO4, 1 mole of PbSO4 will be produced.
Number of moles of PbSO4 produced = 0.0171 moles * (1 mol PbSO4 / 1 mol Na2SO4)
= 0.0171 moles
Finally, we can calculate the mass of PbSO4 produced using its molar mass:
Molar mass of PbSO4 = 207.2 g/mol (atomic mass of Pb) + 32.1 g/mol (atomic mass of S) + (4 * 16.0 g/mol) (4 atoms of O)
= 207.2 g/mol + 32.1 g/mol + 64.0 g/mol
= 303.3 g/mol
Mass of PbSO4 produced = Number of moles of PbSO4 * Molar mass of PbSO4
= 0.0171 moles * 303.3 g/mol
≈ 5.19 grams
Therefore, approximately 5.19 grams of PbSO4 can be produced.
To find the number of grams of PbSO4 produced, we need to determine the limiting reactant in the reaction. The limiting reactant is the one that will be completely consumed and determines the maximum amount of product that can be formed.
To determine the limiting reactant, we can calculate the moles of Na2SO4 and Pb(NO3)2 present in the given volumes and concentrations.
First, convert the volume of each solution to moles using the Molarity (M) formula:
Moles = Volume (L) x Concentration (M)
For Na2SO4:
Moles of Na2SO4 = 0.100 L x 0.171 M = 0.0171 moles
For Pb(NO3)2:
Moles of Pb(NO3)2 = 0.100 L x 0.889 M = 0.0889 moles
Looking at the balanced equation, we can see that the stoichiometric ratio between Na2SO4 and PbSO4 is 1:1. This means that for every mole of Na2SO4, 1 mole of PbSO4 is produced.
Since the stoichiometric ratio is 1:1, the limiting reactant is the one that has the smaller number of moles. In this case, Na2SO4 has fewer moles (0.0171 moles) compared to Pb(NO3)2 (0.0889 moles). Therefore, Na2SO4 is the limiting reactant.
To calculate the number of grams of PbSO4 produced, we need to use the molar mass of PbSO4, which is 303.25 g/mol.
Grams of PbSO4 = Moles of PbSO4 x Molar mass of PbSO4
Since the stoichiometric ratio is 1:1, the number of moles of PbSO4 produced is also 0.0171 moles.
Grams of PbSO4 = 0.0171 moles x 303.25 g/mol = 5.50 grams
Therefore, 5.50 grams of PbSO4 can be produced in this reaction.
mols Na2SO4 = M x L = ?
mols Pb(NO3)2 = M x L = ?
It should be obvious that Na2SO4 is the limiting reagent. 1 mol Na2SO4 produces 1 mol PbSO4; therefore, mols PbSO4 = mols Na2SO4.
Then grams PbSO4 = mols PbSO4 x molar mass PbSO4