Question

date ory Assignment (d) How many moles of barium sulfate were ac- tually formed? mol answer (a) What was the percent yield of barium sulfate?

Answer 1

Molecular equation: K_SO4(aq) + Ba(NO3)2 (aq) - BaSO4 (s) + 2KNO3(aq) Total ionic equation: 2K+ (aq) + SOA? (aq) + Ba?*ceq) + 2NO3 (aq) — BaSO4 () + 2K+ (aq) + 2NO3(aq) Subtract spectator ions: 2K() + SO4?caq) + Bateq) + 2NO, (aq) — BaSO4 (5) + 2K(29) + 2NO3(aq) Net ionic equation: SO4 (aq) + Ba?(aq) — BaSO4 (5)

a) Calculation of the number of moles of the sulfate ion in the 100.0 mL 0.0500 M of the potassium sulfate solution: moles of K, SO = 100.0 mt x 1E/1000 ml x 0.0500 mol/1+ = 0.005 mol KSO4(aq) + 2K+ (aq) +SO4 (ag) The ratio between K, 50, and So, is 1:1 mol of So, = 0.005 mol of K,SO1mol 50,71melk,SO, = 0.005 mol

b) Calculation of the number of moles of the barium ion in the 100.0 mL 0.0700 M of the bariun nitrate solution: moles of Ba(NO3)2 = 100.0 mEx 181000 mE x 0.0700 mol/1+ = 0.007 mol Ba(NO3)2(aq) = Ba²+(aq) + 2NO3 (aq) The ratio between Ba(NO), and Ba** is 1:1 mol of Ba²+ = 0.007 mol of Ba(NO3)2 x 1mol Ba²+71mol Ba(NO3)2 = 0.007 mol

) Calculation of the number of theoretical moles that would be formed of barium Sulfate. SO4 (aq) + Ba*(aq) — BaSO4 (s) Limiting reactant moles of SOL = 0.005 mol moles of Ba2+ = 0.007 mol Number of theoretical moles: 1 mol Baso moles of Baso2 = 0.005 mol so - - 4 mol se = 0.005 mol

d) Calculations of the moles of Baso, formed: mass moles of Baso, =- mw 1.158g moles of Baso. =- = 0.004962 mol 233.38 g/mol e) Calculations of the % yield: % yield = more de moles formed L x 100 theoretical moles % yield = 0.004962 mol - X 100 0.005 mol % yield = 99.24