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Solnce55 [7]
3 years ago
5

Combustion Analysis of an unknown hydrocarbon resulted in the capture of 216.00 g of water vapor and 440.00 g of CO2. The total

mass of the hydrocarbon before combustion was 144 g.
(1. How many moles of carbon dioxide are present after combustion?)

(2. What is the empirical formula for the unknown hydrocarbon?)

(3. How many water molecules does the water vapor trap capture?)

(4. The gram-formula mass for the unknown hydrocarbon was determined to be 72 g/mol. How many moles of the unknown hydrocarbon were present in the original sample?)

(5. infrared spectral analysis determines that the unknown hydrocarbon does NOT contain oxygen. How many moles of O2 are used in combusting the unknown sample?)
Chemistry
1 answer:
LenKa [72]3 years ago
4 0

1: 10 mol CO2

2: C10H24

3: 12 mol H2O

4: 0.5 mol

5: 16 mol O2

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<u>Answer:</u> The enthalpy of the reaction is coming out to be -902 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(4\times \Delta H_f_{(NO(g))})+(6\times \Delta H_f_{(H_2O(g))})]-[(4\times \Delta H_f_{(NH_3(g))})+(5\times \Delta H_f_{(O_2)})]

We are given:

\Delta H_f_{(NO(g))}=91.3kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-45.9kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(4\times (91.3))+(6\times (-241.8))]-[(4\times (-45.9))+(5\times (0))]\\\\\Delta H_{rxn}=-902kJ

Hence, the enthalpy of the reaction is coming out to be -902 kJ.

8 0
4 years ago
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