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    1) A 3.56 g sample of calcium chloride was dissolved in water and reacted with an
    excess of sulfuric acid to form a precipitate of calcium sulfate.
    The percentage yield of calcium sulfate was 83.4%.
    Calculate the mass of calcium sulfate formed.
    Give your answer to an appropriate number of significant figures.

    METHOD =
    CaCl2 moles - (3.56/111.1) = 3.204 x 10-2
    CaSO4 moles - (3.204 x 10-2) x 0.834 = 2.67 x 10-2

    2) The equation for the reaction of ethanal with an alkaline solution of iodine is
    CH3CHO + 3l2 + 4NaOH =CHl3 + HCOONa + 3Nal + 3H2O
    In an experiment using this reaction, the yield of triiodomethane (CHl3) obtained by a
    student was 83.2%.
    Calculate the minimum mass of iodine that this student would have used to form 10.0 g of
    triiodomethane

    If i use the same method as above, answer would be
    CHI3 moles = (10/393.7) = 0.0254
    0.0254 x 0.832 = 0.02113
    3I2 moles = 3 x 0.02113 = 0.06339
    3I2 mass = 0.06339 x (126.9x2) = 16.1 g
    but this is wrong and the M.S. uses a different method....why??????????

    p.s. sorry for long post
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    Question 1, you still need to find the mass by multiplying by the Mr of CaSO4.

    Question 2, you went wrong at the step
    "0.0254 x 0.832 = 0.02113"

    The correct step is 0.0254/0.832.

    if yield was 100%, number of moles of CHI3 produced = 10/393.7 and number of moles of I2 = 3 x 10/393.7
    however, yield is only 83.2%. You need to use more of I2.
    so in fact the number of moles of I2 needed = (3 x 10/393.7)/0.832
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    (Original post by BobbJo)
    Question 1, you still need to find the mass by multiplying by the Mr of CaSO4.

    Question 2, you went wrong at the step
    "0.0254 x 0.832 = 0.02113"

    The correct step is 0.0254/0.832.

    if yield was 100%, number of moles of CHI3 produced = 10/393.7 and number of moles of I2 = 3 x 10/393.7
    however, yield is only 83.2%. You need to use more of I2.
    so in fact the number of moles of I2 needed = (3 x 10/393.7)/0.832
    thanks for your reply!

    so why for the first question do you have to multiply by the percentage but for q2, you have to divide by the percentage yield
    btw i forgot to type in the last step of q1 so
    mass = (2.672 x 10-2) x 136.2 = 3.6g
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    (Original post by jonboi1235)
    thanks for your reply!

    so why for the first question do you have to multiply by the percentage but for q2, you have to divide by the percentage yield
    btw i forgot to type in the last step of q1 so
    mass = (2.672 x 10-2) x 136.2 = 3.6g
    if it was 100% percentage yield,
    you would need to use 3 x 10/393.7 moles of iodine to get 10 g of CHI3

    however we are only getting 83.2% yield,
    so the 3 x 10/393.7 moles of iodine will only yield 10 x 0.832 g of CHI3
    hence moles of iodine required to get 10 g of CHI3 is (3 x 10/393.7)/0.832
 
 
 

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