You just need to know this one.

2. How many hydrogen atoms are there in ethanol?

Hydrogen Atoms

Six. C₂H₅OH. Some people forget the one part of the hydroxy group.

3. Given the below information, find the molar heat of combustion for ethanol.

    Initial Water Temp: 23^oC

    Final Water Temp: 40^oC

    Initial Mass of Spirit Burner: 207.60 g

    Final Mass of Spirit Burner: 206.61 g

    Mass of water used: 200 g

kJ mol^-1 (Whole Number)

Firstly, we need to work out the energy liberated using the formula:

ΔH = -100 x 4.18 x 17

ΔH = -14212 J

Now we need to work out how many grams of ethanol we used:

Δm = 207.60 - 206.61

Δm = 0.99 grams

Now we simply multiply the amount of heat liberated to convert it to a per mole value.

ΔH = 14212 J per 0.99 grams

ΔH = 14212 x (46.068 / 0.99)

ΔH = 661 331.7 J mol^-1

ΔH = 661 kJ mol^-1

4. How many grams of ethanol would it take to produce 10kJ of energy given:

    Ethanol's Molar Heat of Combustion: 1360 kJ mol^-1

    The temperature rose 10^oK.

grams (Two decimal places)

The amount the temperature rose is a red herring. It is not important. To do this question all we need to do is a simple calculation.

Moles needed to produce 10 kJ = 10 / 1360

Moles needed to produce 10 kJ = 0.007352941176

Multiply this by the molar mass of ethanol to get the mass required:

m = 0.00735294 x 46.068

m = 0.34 grams

5. A student undertook an experiment to determine the molar heat of combustion for 1-Butanol. Given the following information, how much water did the student use?

    Temperature Rise: 15^oK

    Energy liberated: 23.9 kJ

mL (Whole Number)

Once again, we use the formula:

23900 = m x 4.18 x 15

m = 381 grams

Water = 381 mL

grams (Three decimal places)

Ethanol Formula: C₂H₆O

Molar Mass = (2 x Carbon) + (6 x Hydrogen) + Oxygen

Molar Mass = (2 x 12.01) + (6 x 1.008) + 16

Molar Mass = 46.068 grams

7. The following equation is NOT balanced. Balance it. How many moles of ethanol would be produced if three moles of glucose underwent fermentation?

     C₆H₁₂O₆  -->  CH₃CH₂OH  +  CO₂

Moles (Whole Number)

Balanced Equation:

C₆H₁₂O₆  -->  2CH₃CH₂OH  +  2CO₂

Therefore the ratios are:

   1        :           2             :    2

So if three moles of glucose underwent fermentation, twice that would be produced. Hence, six moles.

Moles (One decimal place)

Moles = Mass  /  Molar Mass

Moles = 300 / 46.068

Moles = 6.5

V (Two decimal places)

Looking at your table of standard potentials you will see:

 

Zn²⁺  +  2e^-  -->  Zn_(s)       -0.76 V

Cu⁺  +  e^-  -->  Cu_(s)           0.52 V

 

However, the above equations are reduction half equations. Since zinc is at the anode, it undertakes oxidation and we therefore need to switch its equation around.

 

Zn_(s)  -->  Zn²⁺  +  2e^-     0.76 V

 

Now we add the potentials of the two half equations together.

E = 0.76 + 0.52

E = 1.28 V

^oC (Nearest 100 degrees)

Just need to know this one.