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
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
What mass of barium sulfate (233 g/mol) is produced when 125 mL of a 0.150 M...
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