Mass calculation and Theoretical yield help?

Calculate the mass of potassium dichromate (K2Cr2O7) required to produce 1.0 g of iodine (I2) according to the equation below:

K2Cr2O7 + 6NaI + 7H2SO4 à Cr2 (SO4)3 + 3I2 + 7H2O + 3Na2SO4 + K2SO4

I already balanced it

two--Calculate the theoretical yield percentage of a reaction between sodium and water, if 25.0 g of sodium react with excess water and if 15.5 L of gaseous hydrogen are collected at NTP conditions

2 Answers

  • 1 decade ago
    Favorite Answer

    K2Cr2O7 + 6NaI + 7H2SO4 --> Cr2(SO4)3 + 3I2 + 7H2O + 3Na2SO4 + K2SO4

    Part1: You want to produce 1g of I2. The molecular mass of I2 is 2*126.90447g/mol = 253.80894/mol

    1g I2 * (1mol / 253.80894g) = 0.00393997154mol I2

    Now, using the balanced chemical equation, you can convert moles of I2 to moles of K2Cr2O7

    0.00393997154mol I2 * (1mol K2Cr2O7 / 3mol I2) = 0.00131332385mol K2Cr2O7

    Now, convert the moles of K2Cr2O7 back to mass, using its molecular mass 294.1846 g/mol.

    0.00131332385mol K2Cr2O7 * (294.1846 g/mol) = 0.386359651g K2Cr2O4.

    Part2: I'm going to assume the chemical reaction is Na + H2O --> NaOH + H2. Balancing this gives

    2Na + 2H2O --> 2NaOH + H2

    Since you have excess water, sodium is the limiting reagent. Its molecular mass is 22.99g/mol.

    25g Na * (1mol / 22.99g) = 1.08742932mol Na.

    Converting this to moles of Hydrogen gas using the equation, we get

    1.08742932mol Na * (1mol H2 / 2 mol Na) = 0.54371466mol H2.

    This is the theoretical yield.

    Keeping this in mind for later, let's get the actual yield. Use the following equation, using the constant R=0.0821 and keeping in mind that NTP (Normal Temperature and Pressure) is equal to 293K and 1atm, respectively:


    (1)(15.5) = n(0.0821)(293)

    n = 15.5 / (0.0821 * 293) = 0.644348647mol H2

    So, theoretical percentage yield refers to actual yield over theoretical yield.

    If we crunch the numbers, TPY = (0.644348647/0.54371466) * 100% = 118.508603% yield.

    Source(s): It is strange that the percent yield is over 100%. I could have made an error, but I did my best to check my work.
  • 4 years ago

    because the actually yield is received by an attempt (as adverse to calculated from the periodic table like the theoretical yield) you likely measured some thing incorrect, or used too a lot or a substance or your stability would were off. it truly is likely why your actually yield became off.

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