6. What is the molecular formula for this compound? The compound's empirical formula and
1 answer:
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Answer: 824.6 g of NaCl are produced from 500.0 g of chlorine.
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of produce = 2 moles of
Thus 7.04 moles of will produce=
of
Mass of
Thus 824.6 g of NaCl are produced from 500.0 g of chlorine.
For this reaction to proceed, the following bond breaking should occur:
*one C-H bond
* one Cl-Cl bond
After, the following bond formations should occur:
*one C-Cl bond
*one H-Cl bond
Now, add the bond energies for the respective bond energies which can be found in the attached picture. For bond formations, energy is negative. For bond breaking, energy is positive.
ΔHrxn = (1)(413) + (1)(242) + 1(-328) + 1(-431) = <em>-104 kJ</em>
*one C-H bond
* one Cl-Cl bond
After, the following bond formations should occur:
*one C-Cl bond
*one H-Cl bond
Now, add the bond energies for the respective bond energies which can be found in the attached picture. For bond formations, energy is negative. For bond breaking, energy is positive.
ΔHrxn = (1)(413) + (1)(242) + 1(-328) + 1(-431) = <em>-104 kJ</em>