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24 March, 17:04

3. Predict the mass of Na2S2O3that could be produced from 38.2 g of Na2S and unlimited quantities of the other reactants, based on the following unbalanced equation:

Na2CO3 + Na2S + SO2--> Na2S2O3 + CO2

a. If the procedure produced only a 68.5 % reaction yield, how many grams of Na2S2O3would you expect to obtain?

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Answers (2)
  1. 24 March, 18:27
    0
    We would obtain 79.4 grams of Na2S2O3

    Explanation:

    Step 1: Data given

    Mass of Na2S = 38.2 grams

    Molar mass Na2S = 78.05 g/mol

    Molar mass Na2S2O3 = 158.11 g/mol

    Reaction yield = 68.5 %

    Step 2: The balanced equation

    2Na2S + Na2CO3 + 4SO2 → 3Na2S2O3 + CO2

    Step 3: Calculate moles Na2S

    Moles Na2S = mass Na2S / molar mass Na2S

    Moles Na2S = 38.2 grams / 78.05 g/mol

    Moles Na2S = 0.489 moles

    Step 4: Calculate moles Na2S2O3

    For 2 moles Na2CO3 we need 1 mol Na2S and 4 moles SO2 to produce 3 mol Na2S2O3 and 1 mol CO2

    For 0.489 moles Na2S we'll have 3/2 * 0.489 = 0.7335 moles Na2S2O3

    If 0.7335 moles is produced the % yield is 100 %

    For a 68.5 % yield, 0.7335 * 0.685 = 0.5024 moles

    Step 5: Calculate mass Na2S2O3

    Mass Na2S2O3 = 0.5024 moles * 15.11 g/mol

    Mass Na2S2O3 = 79.4 grams

    For a 100 % yield, we would obtain 115.97 grams of Na2S2O3

    For a 68.5 % yield we would obtain 79.4 grams of Na2S2O3
  2. 24 March, 18:45
    0
    1. 116.07g

    2. 79.51g

    Explanation:

    1. First, let us write a balanced equation for the reaction. This is illustrated below:

    Na2CO3 + 2Na2S + 4SO2 - > 3Na2S2O3 + CO2

    Next, let us calculate the mass of Na2S that reacted and the mass of Na2S2O3 produced from the balanced equation.

    This is illustrated below:

    Molar Mass of Na2S = (23x2) + 32 = 46 + 32 = 78g/mol

    Mass of Na2S from the balanced equation = 2 x 78 = 156g

    Molar Mass of Na2S2O3 = (23x2) + (32x2) + (16x3) = 46 + 64 + 48 = 158g/mol

    Mass of Na2S2O3 from the balanced equation = 3 x 158 = 474g

    From the balanced equation above,

    156g of Na2S produced 474g of Na2S2O3.

    Therefore, 38.2g of Na2S will produce = (38.2x474) / 156 = 116.07g of Na2S2O3

    From the calculations made above, 38.2g of Na2S will produce 116.07g of Na2S2O3.

    2. The second part of the question shows that the percentage yield is 68.5%

    Now, the grams we are expecting to have, talks about the experimental yield.

    From the calculations made above, the theoretical yield is 116.07g. The experimental yield can be calculated for as shown below:

    %yield = Experimental yield / Theoretical yield x100

    68.5% = Experimental yield / 116.07

    68.5/100 = Experimental yield / 116.07

    Cross multiply to express in linear form as shown below:

    100 x Experimental yield = 68.5x116.07

    Divide both side by 100

    Experimental yield = (68.5x116.07) / 100

    Experimental yield = 79.51g

    Therefore, we are expected to obtain 79.51g of Na2S2O3 since the percentage yield is 68.5%
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