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

Question 1 of 3

You need to produce 28 litres of Methane (measured at 0°C and 760 mm Hg). If Hydrochloric Acid is present in excess, what is the MINIMUM amount in grams of Aluminium Carbide that you must use, assuming a percentage yield of 100 %. Answer to 1 decimal place. The reaction is as below:

Al4C3 + 12HCl ---------» 3CH4 + 4AlCl3
Type in your answer and PRESS ENTER.
grams
You would need at least 60.0 grams of Aluminium Carbide.
  1. Find the number of moles of Methane that you need to make.
    Moles of Methane = Volume/22.4 moles
    Moles of Methane = 28 / 22.4 = 1.25 moles
  2. Find how many moles of Aluminium Carbide you need.
    From the balanced equation you can see that:
    1 mole of Methane is produced from 1/3 moles of Aluminium Carbide. Thus:
    1.25 moles of Methane are produced from 1.25 x 1/3 moles of Aluminium Carbide.
    1.25 moles of Methane are produced from 0.41667 moles of Aluminium Carbide.
  3. Find the mass of Aluminium Carbide needed.
    Mass of Aluminium Carbide needed = Moles x RMM           RMM = 144.0
    Mass of Aluminium Carbide needed = 0.41667 x 144.0 g = 60 g = 60.0 g 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.

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

Question 2 of 3

You need to prepare 37.3 grams of Potassium Chloride. What is the minimum volume (in millilitres) of 2.000 M Potassium Hydroxide you would need if you were preparing Potassium Chloride using the following reaction:

KOH + HCl ---------» KCl + H2O
Type in your answer and PRESS ENTER.
ml
The minimum volume of Potassium Hydroxide is 250 millilitres.
  1. Find the number of moles of Potassium Chloride to be produced:
    Moles = Mass / RMM                            RMM = 74.6
    Moles of Potassium Chloride = 37.3 / 74.6 = 0.5 moles
  2. Find how many moles of Potassium Hydroxide are needed.
    From the balanced equation you can see that:
    1 mole of Potassium Chloride is produced from 1 mole of Potassium Hydroxide. Thus:
    0.5 moles of Potassium Chloride are produced from 0.5 moles of Potassium Hydroxide.
    So, 0.5 moles of Potassium Hydroxide are needed.
  3. Find the volume of 2.000 M Potassium Hydroxide needed.
    Molarity = Moles / Volume (in litres)               Volume (in litres) = Moles / Molarity
    Volume of Potassium Hydroxide = 0.5 / 2.000 litres = 0.25 litres
    Volume of Potassium Hydroxide needed = 250 ml

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

In the following reaction:
C2H5OH + HBr ---------» C2H5Br + H2O

How many grams of Ethanol would you need in order to prepare 36.3 grams of Ethyl Bromide, assuming a 100 % percentage yield? Answer to 2 decimal places.

Type in your answer and PRESS ENTER.
grams
You would need 15.33 grams of Ethanol.
  1. Find how many moles of Ethyl Bromide are to be produced.
    Moles of Ethyl Bromide = Mass / RMM                               RMM = 108.9
    Moles of Ethyl Bromide = 36.3 / 108.9 = 1/3 moles
  2. Find how many moles of Ethanol are needed.
    From the balanced equation you can see that:
    1 mole of Ethyl Bromide is produced from 1 mole of Ethanol. Thus:
    1/3 moles of Ethyl Bromide are produced from 1/3 mole of Ethanol.
    Thus, 1/3 moles of Ethanol are needed.
  3. Find the mass of Ethanol needed.
    Mass of Ethanol = Moles / RMM                                          RMM = 46.0
    Mass of Ethanol = 1/3 x 46.0 g = 15.333 g
    Mass of Ethanol = 15.33 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