Question

What mass of barium sulfate (233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 25 mL of a 0.175 M solution of iron(III) sulfate

Answer 1

3BaCl2+ Fe2(SO4)3?3BaSO4+ 2FeCl3

0.0125 L x 0.150 M = 0.001875 moles barium chloride

0.0025 L x 0.175 M = 0.0004375 moles iron(iii) sulphate

3 moles of BaCl2 reacts with 1 mole of iron(iii) sulphate

=> 0.001875 moles of BaCl2 reacts with (1/3) * 0.001875 = 0.000625 moles of iron(iii) sulphate

so, here the limiting agent is iron(iii) sulphate

so, the 1 mole of iron(iii) sulphate produces 3 moles of barium sulphate

so, 0.0004375 moles iron(iii) sulphate produces 3*0.0004375 = 0.0013125 moles of barium sulphate

as the molecular weight of barium sulphate = 233 g/mole

so the amount of barium sulphate produce = 233 * 0.0013125 = 0.3058 g

Answer 2

Given that Concentration of Bac, 0.150M Volume of BaC,-125mL Concentration of Fe SO0.175M Volume of Fe.SO4-25mL Molar mass of BaSO4 233 mol Consider the reaction, Here, 3 moles of BaCl, reacts with one mole of Fe,SO4 and gives 3 moles of BaSO4 and 2 moles of FeCl, Calculate the number of moles of BaC Moles of BaCl -Concentration of BaCl,xVolume of BaCl, 1E 1000 paL mol 0.0187 mol Calculate the number of moles of Fe,SO Moles of Fe,SO4-Concentration of Fe,SO4x Volume of Fe,SO,4 mol 1000 paL 0.0044mol According to the above reaction, 3 moles of BaCl, require with one mole of Fe,SO,4

Hence, for 0.0187 moles of BaCl,, the number of moles of Fe,SO required would be Moles of Fe,SO,-0.0187molx- -0.00623 mol However, here, only 0.0044 mol of Fe,SO, is present. Therefore, Fe SO, is the limiting reagent in the reaction and BaSO, will be formed from the available amount of Fe,SO In the reaction, one mole of Fe,So, is used to produce 3 Calculate the number of moles of BaSO Moles of Baso, 3*Moles of Fe,SO, moles of BaSO, 3x0.0044mol 0.0132mol Calculate the mass of BaSO. Massof Baso,-Moles of Baso, Molar massof BaSo, -0.0132mølx233 moi 3.075g Hence, 3.075gof BaSO is produced in the given reaction.