Question

what is the volume of 0.200 M sulfiric acid needed to react with excess of iron (III) oxide to produce 4.00 g of iron (III) sulfate if the percent yield for the reaction is 50%?

1 Fe2O3 + 3H2SO4 = 1 Fe2(SO4)3 + 1 H20

Answer 1

Molar mass of iron iii sulphate = 399.9 g/mol Calculate the moles of iron sulphate produced: Iron iii sulphate = given mass _ 4. mole molar mass 399.9" - - - Then moles of sulphuric acid used = 3 x 2x Iron iii sulphate (since 50% yield) = 6x200 . moles Calculate the volume of sulphuric acid used as follows: Volume = moles conc. 4 6x Volume = _ 399.9 0.200 Volume=0.3 L Volume = 300 mL Please give good rating.