Solutions of sulfuric acid and lead acetate react to form solid lead sulfate and a solution of acetic acid.5.80 of sulfuric acid and 5.80 of lead acetate are mixed.
Calculate the number of grams of acetic acid present in the mixture after the reaction is complete.
The reaction is as followed:
H2SO4 + Pb(C2H3O2)2----> PbSO4 +2(C2H4O2)
However, unless you provide units for your reactants, there is no way to start this problem.
i don't know i need help
To calculate the number of grams of acetic acid present in the mixture after the reaction is complete, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed, which will determine the maximum amount of product that can be formed.
To find the limiting reactant, we need to calculate the number of moles of sulfuric acid and lead acetate separately. The molar mass of sulfuric acid (H2SO4) is 98.09 g/mol, and the molar mass of lead acetate (Pb(CH3COO)2) is 325.29 g/mol.
First, let's calculate the number of moles of sulfuric acid:
moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4
moles of H2SO4 = 5.80 g / 98.09 g/mol
moles of H2SO4 ≈ 0.059 mol
Now, let's calculate the number of moles of lead acetate:
moles of Pb(CH3COO)2 = mass of Pb(CH3COO)2 / molar mass of Pb(CH3COO)2
moles of Pb(CH3COO)2 = 5.80 g / 325.29 g/mol
moles of Pb(CH3COO)2 ≈ 0.018 mol
To determine the limiting reactant, we compare the moles of each reactant to the stoichiometry of the balanced equation. From the balanced equation:
1 mole of H2SO4 reacts with 1 mole of Pb(CH3COO)2 to form 1 mole of acetic acid.
Since the ratio of moles between H2SO4 and Pb(CH3COO)2 is 1:1, we can see that the moles of H2SO4 are greater than the moles of Pb(CH3COO)2. Therefore, the limiting reactant is Pb(CH3COO)2.
Now, we can calculate the moles of acetic acid produced using the limiting reactant:
moles of acetic acid = moles of Pb(CH3COO)2
moles of acetic acid ≈ 0.018 mol
Finally, we convert the moles of acetic acid to grams using the molar mass of acetic acid (CH3COOH), which is 60.05 g/mol:
mass of acetic acid = moles of acetic acid x molar mass of acetic acid
mass of acetic acid ≈ 0.018 mol x 60.05 g/mol
mass of acetic acid ≈ 1.083 g
Therefore, the number of grams of acetic acid present in the mixture after the reaction is complete is approximately 1.083 grams.