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- Homework Statement
- Calculate the heat Kcal measured at 25.0°C and 1 atm corresponding to the complete combustion of 4.00 L of Benzene (C6H6 d= 0.780 6/mL) when liquid water forms.

- Relevant Equations
- thermochemistry

The balanced reaction wil be :

##2C_6H_{6(l)}+15O_{2(g)}->12CO_{2(g)}+6H_2O_(l)##

in order to compute the the standard enthalpy of reaction :

##\Delta H°_{f} H_2O_(l)= -285,8 \frac {KJ}{mol}##;

##\Delta H°_{f} CO_{2(g)}= -393,5 \frac {KJ}{mol}##;

##\Delta H°_{f} C_6H_{6(l)}=49,04 \frac {KJ}{mol}##;

##\Delta H°_{R}= \sum \Delta H°_{f} \text{ products} -\sum \Delta H°_{f} \text{ reagents}= [(-285,8 \frac {KJ}{mol}*6\text{ mol})+(-393,5 \frac {KJ}{mol}*12\text{ mol})]-(49,04\frac {KJ}{mol}*2\text{ mol})=-6531,26 KJ##

converting L to mL:

4,00L=4000 mL

The grams of ##2C_6H_{6(l)}##:

4000 mL *0,789 ##\frac g {mol}##= 3120 g;

Moles of ##2C_6H_{6(l)}##:

##\frac {3120 g}{6*12,01\frac g {mol}+6*1,01\frac g {mol} }##= 39,9 mol;

By proportion the moles of liquid water formed will be :

mol ##H_2O_(l)##=## \frac{6 \text{ mol } H_2O_(l)*39,9\text{ mol } 2C_6H_{6(l)}}{2 \text{ mol }2C_6H_{6(l)}}##= 120 mol

The heat , by proportionality :

##\frac{120 \text{ mol } *(-6531,26 KJ)}{6 \text{ mol }}##= -131*10^3 KJ = -3,13*10^4 Kcal

Is it all correct in your opinion?

the book says 3,03 *10 ^4 Kcal

the allowed error is 3% by the professor, where am i wrong?

##2C_6H_{6(l)}+15O_{2(g)}->12CO_{2(g)}+6H_2O_(l)##

in order to compute the the standard enthalpy of reaction :

##\Delta H°_{f} H_2O_(l)= -285,8 \frac {KJ}{mol}##;

##\Delta H°_{f} CO_{2(g)}= -393,5 \frac {KJ}{mol}##;

##\Delta H°_{f} C_6H_{6(l)}=49,04 \frac {KJ}{mol}##;

##\Delta H°_{R}= \sum \Delta H°_{f} \text{ products} -\sum \Delta H°_{f} \text{ reagents}= [(-285,8 \frac {KJ}{mol}*6\text{ mol})+(-393,5 \frac {KJ}{mol}*12\text{ mol})]-(49,04\frac {KJ}{mol}*2\text{ mol})=-6531,26 KJ##

converting L to mL:

4,00L=4000 mL

The grams of ##2C_6H_{6(l)}##:

4000 mL *0,789 ##\frac g {mol}##= 3120 g;

Moles of ##2C_6H_{6(l)}##:

##\frac {3120 g}{6*12,01\frac g {mol}+6*1,01\frac g {mol} }##= 39,9 mol;

By proportion the moles of liquid water formed will be :

mol ##H_2O_(l)##=## \frac{6 \text{ mol } H_2O_(l)*39,9\text{ mol } 2C_6H_{6(l)}}{2 \text{ mol }2C_6H_{6(l)}}##= 120 mol

The heat , by proportionality :

##\frac{120 \text{ mol } *(-6531,26 KJ)}{6 \text{ mol }}##= -131*10^3 KJ = -3,13*10^4 Kcal

Is it all correct in your opinion?

the book says 3,03 *10 ^4 Kcal

the allowed error is 3% by the professor, where am i wrong?