Question

6. (6.0 pts) If 15.00 g of FeBr₃ is combined with 8.50 g of H₂SO₄, how many grams of Fe₂(SO₄)₃ will be produced?

7. (2.0 pts) If only 6.25 g of Fe₂(SO₄)₃ is actually produced, what is the percent yield?

Answer 1

6.

L the elements frounded up to nearest Molecular weights of the elements (rou integer): Fe = 56 asmole ; Br = 80 H = 1 g/mole ; S = 32 g/mole 0 = 16 g/mole Now, molecular weights cular, weights of of the compounds mentioned i. the Fe Brz § 56 + (3 x 80 ) = 296 g/mole. H2SO4 F (2 x 1) + 32 +(4x16) = 98 g/mole Fez (504), = (2 x 56) + 3 x( 32 + (4x169) – 112 + (3x96) The reaction for the = 400 g/mole problem :- = Fe₂(SO4)3 + 6HB (Balanced equation. 2Fe Brz + 3 H₂SO4 we have, fe Bra 15 296 mole 8.5 Therefore, 2 moles Fear, and 3 moles H₂SO4 have to combine to form 1 mole Fe₂(SO4). 15.009 296 g/mole H₂SO4 = 8.50g = 8.50 a To 198 g/mole since, 2 moles Febrz gives a mole Fe₂(SO4)3 then, 2 " " will give x 15 mole Fe2(SO4)3. = 296 x 400 = 10-14 g fez(504) - mole 98 IS 98 Similarly, if we calculate for H₂50g :- 3 moles! H2509 = 1 mole fe2804)3 8.5 = x š x 400 g * 11.56 g fez(s047 product always form with respect to the reactant, which is present in lowest amount ie here with respect to FeBrz. Because, H₂SO4 alone casinot give any product? the although it is present in greater amount than fer produced. Therefore, 10.14 grams of Fe₂(SO4)3 will be This is the theoretical yield in grams.]

7. Now, actual yield in grams (given) = 6.25 g.

As we know, percentage yield  

actual yield produced theoretical yield x 100 %

6.25 8 x 100 % = 20 10.14 g

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= 61.64%