For the next question, click on the Next Question button.

Question 1 of 3

What is the theoretical volume of Methane in litres (measured at 0°C and 760 mm Hg) produced from 23.9 grams of Trichloromethane in the following reaction:

CHCl3 + 3Zn + 3H2O ---------» CH4 + 3Zn(OH)Cl
Type in your answer and PRESS ENTER.
litres
The theoretical yield of Methane is 4.48 litres.
  1. Find the number of moles of Trichloromethane used.
    Moles of Trichloromethane = Mass / RMM                            RMM = 119.5
    Moles of Trichloromethane = 23.9/119.5 = 0.2 moles
  2. Find how many MOLES of Methane are produced.
    From the balanced equation you can see that:
    1 moles of Trichloromethane would produce 1 mole of Methane. Thus:
    0.2 moles of Trichloromethane would produce 0.2 moles of Methane.
  3. Find the volume of Methane produced - as measured at 0°C and 760 mm Hg.
    1 moles of Methane would have a volume of 22.4 litres. Thus:
    0.2 moles of Methane would have a volume of 0.2 x 22.4 litres = 4.48 litres.

No, that is not the correct answer.

Have another go: just type in your new answer and press Enter.
If you continue to have problems, have a look at the answer.

For the next question, click on the Next Question button.

Question 2 of 3

What is the theoretical yield in milligrams of Sodium Sulphate in the following reaction, involving 25 ml of 0.1000M Sodium Hydroxide and 30 ml of 0.1000 M Sulphuric Acid?
Answer to 1 decimal place.

2NaOH + H2SO4 ---------» Na2SO4 + 2H2O
Type in your answer and PRESS ENTER.
milligrams
The theoretical yield of Sodium Sulphate is 177.5 milligrams.
  1. Find the number of moles of each of the reactants.
    Molarity = Moles/Volume (in litres); thus: Moles = Molarity x Volume
    Moles of Sodium Hydroxide    = 0.1000 x 0.025 moles = 0.0025 moles
    Moles of Sulphuric Acid          = 0.1000 x 0.030 moles = 0.003 moles
  2. Find which of Sodium Hydroxide and Sulphuric Acid is the limiting reactant.
    From the balanced equation, you can see that:
    2 moles of Sodium Hydroxide would produce 1 mole of Sodium Sulphate. Thus:
    0.0025 moles of Sodium Hydroxide would produce 0.00125 moles of the Sulphate.
    1 mole of Sulphuric Acid would produce 1 mole of Sodium Sulphate
    Thus 0.003 moles of Sulphuric Acid would produce 0.003 moles of Sodium Sulphate.
  3. Find the mass of Sodium Sulphate produced.
    Mass = Moles x RMM                      RMM = 142.0
    Mass = 0.00125 x 142 g = 0.1775 g = 177.5 milligrams to 1 dec pl

No, that is not the correct answer.

Have another go: just type in your new answer and press Enter.
If you continue to have problems, have a look at the answer.

Question 3 of 3

What is the theoretical yield in grams of Ethyl Iodide in the following reaction, involving 69 grams of Ethanol? Answer to 2 decimal place.

2P + 3I2 + 6C2H5OH ---------» 6C2H5I
Type in your answer and PRESS ENTER.
grams
The theoretical yield of Ethyl Iodide is 233.85 grams.
  1. Find the number of moles of Ethanol used.
    Moles of Ethanol = Mass/RMM                    RMM = 46.0
    Moles of Ethanol = 69/46.0 = 1.5 moles
  2. Find the number of moles of Ethyl Iodide produced. Using the balanced equation:
    6 moles of Ethanol would produce 6 moles of Ethyl Iodide. Thus:
    1 mole of Ethanol would produce 1 mole of Ethyl Iodide. Thus:
    1.5 moles of Ethanol would produce 1.5 moles of Ethyl Iodide.
  3. Find the MASS of Ethyl Iodide produced.
    Mass of Ethyl Iodide = Moles x RMM           RMM = 155.9
    Mass of Ethyl Iodide = 1.5 x 155.9 g
    Mass of Ethyl Iodide = 233.85 g to 2 dec pl

No, that is not the correct answer.

Have another go: just type in your new answer and press Enter.
If you continue to have problems, have a look at the answer.

Click to Continue