Question

2. A precipitate is formed when 2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 L of 0.0664 M N^2SO4 based upon the equation below Ba(OH)2(aq)Na2SO4(aq) > BaSO4(s) + 2 N2OH(aq) What is the mass of the precipitate formed from the Ba(OH)2? (Show all work) What is the mass of the prccipitatc formcd from thc Na2SO4? (Show all work) What is the limiting reactant? What is the theoretical yield of the precipitate?

Answer 1

The given equation is --Ba (061)z fegy + Na, SO4 604) > Bason est + 2NaOH(ate According to this balanced equation, - mole for Ba (et) Greacts with moleen of Na, soynto, peroduce I molen to Basove olar we can calculate number of moles of the reactants from their concentrations Since Tholarity = Number of moles of solute volume of solution in L So Neuf moles - fonolarity) (volume of of selute Thus, o Number of moles (0.0820M) (2.276) ba ( ону, — = (0.0820 mollux (2:2747 = 0.186 moles Also, Number of moles -(0.0664 mol 1%) x (3.064) of Na, soy = 0.203 moles

Since I mole of Ba(OH) produces I mole of precipitates of Basen then o. 186 males of Bacou will produce 0:18.6._molesef Ba 304 As Moles - Mass holar Mars so Mass = Moles & Molar Mass OMolor Mase basov » 232-33 amel So mass of precipitates of Basou formed from 0.156 moles of Balon) v e of moles of basoutě (Molas Mods of Basou) = 0.186 mol x 233.38 g/mol = 43.4 g. Thus mass of precipate formed from Ba(OH)2 = pp.ug 6) As I mole of Nasou producent mole of precipitastes ok Basou So 0.203 moles of Nasou should produce - 0.203 moles of Basou precipitates. Mass of Bason = 0.203 mot x 233.38 g/ mot 47.4g- Thus mass of precipitate formed from -Naz sou-funug

c) As I mole of Ba(OH) reacts completely with mole of Na,sou then 0.203 moles of Na,sou will react completely with 0.203 moles of Ba(OH)₂ but there are only 0.186 moles of Ba(OH) present. This Bacot) is the limiting reactant as its amount is limited and all of 0.186 moles of Balat) will be consumed by reaction with 0.186 moles of Nasou Whereas Na, sou will remain an the reaction mixture as its amount is in excess Nasou that would be left over unreacted = 0.203-0.186 moles - = 0.07 moles d Theoretical Yield is always calculated with respect to the limiting reactant as once the limiting reactant ses consumed the reaction will not be carried out further Ds already worked out in part (a), 0. 186 moles of Ba(OH)₂ (the limiting reactant) would produce 0.186 moles of precipitates of Bason that is 3. часоѕ04
Thus theoretical yield of the precipitate is 43.4 g.