A 1.009-g mixture of the solid salts Na2SO4 (molar mass=142.04g/mol) and Pb(NO3)2 (molar mass=331.20g/mol). The precipitate was filtered and dried, and it mass was determined to be 0.471 g. The limiting reactant was determined to be Na2SO4.
a. write the molecular form of the equation for the reaction?
b. write the net ionic equation for the reaction?
c. How many moles and grams of Na2SO4 are in the reaction mixture?
d. How many moles and grams of Pb(NO3)2 reacted in the reaction mixture?
e. what is the percent by mass of each salt in the mixture?
I'll be glad to help you through this if you explain what you don't understand. I can't believe you can't do ANY of this. But I'm not in to doing your lab workup for you.
To answer these questions, we need to analyze the given information and apply stoichiometry calculations. Let's go step by step:
a. To write the molecular form of the equation, we need to find the balanced chemical equation for the reaction between Na2SO4 and Pb(NO3)2. Since we know that Na2SO4 is the limiting reactant, it fully reacts with Pb(NO3)2 in a 1:1 molar ratio. The balanced equation is:
2 Na2SO4 + Pb(NO3)2 → 2 NaNO3 + PbSO4
b. To write the net ionic equation, we need to consider only the ions that participate in the reaction. The soluble salts NaNO3 and Pb(NO3)2 dissociate into their constituent ions in water. The net ionic equation is:
2 Na+ + SO4^2- + Pb^2+ → 2 Na+ + 2 NO3- + PbSO4
c. We are given the mass of the mixture and the limiting reactant, Na2SO4. To find the number of moles of Na2SO4 in the mixture, we divide its mass by its molar mass:
moles of Na2SO4 = mass of Na2SO4 / molar mass of Na2SO4
= 1.009 g / 142.04 g/mol
≈ 0.0071 mol
To find the mass of Na2SO4, we can use its molar mass:
mass of Na2SO4 = moles of Na2SO4 * molar mass of Na2SO4
= 0.0071 mol * 142.04 g/mol
≈ 1.0084 g
d. Since Na2SO4 is the limiting reactant, it reacts completely. Therefore, no Pb(NO3)2 should react. However, we can find the moles of Pb(NO3)2 present in the mixture before the reaction occurred:
moles of Pb(NO3)2 = mass of Pb(NO3)2 / molar mass of Pb(NO3)2
= (total mass - mass of Na2SO4) / molar mass of Pb(NO3)2
= (1.009 g - 1.0084 g) / 331.20 g/mol
≈ 0.0018 mol
To find the mass of Pb(NO3)2 that reacted, we can use its molar mass and subtract it from the initial mass:
mass of Pb(NO3)2 reacted = (moles of Pb(NO3)2 * molar mass of Pb(NO3)2) - mass of Pb(NO3)2
= (0.0018 mol * 331.20 g/mol) - 1.009 g
≈ 0.619 g
e. To find the percent by mass of each salt in the mixture, we need to divide the mass of each salt by the total mass of the mixture and multiply by 100%.
Percent by mass of Na2SO4 = (mass of Na2SO4 / total mass of mixture) * 100%
= (1.0084 g / 1.009 g) * 100%
= 99.91%
Percent by mass of Pb(NO3)2 = (mass of Pb(NO3)2 / total mass of mixture) * 100%
= (0.619 g / 1.009 g) * 100%
= 61.30%
Therefore, the percent by mass of Na2SO4 in the mixture is approximately 99.91%, and the percent by mass of Pb(NO3)2 is approximately 61.30%.