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larisa86 [58]
3 years ago
6

A laboratory manual gave precise instructions for carrying out the reaction described by the chemical equation, C2H6O + O2 C2H4O

2 + H2O. It stated that if you bubble oxygen gas through a solution containing the reactant for 24 hours then the yield of C2H4O2 would be 7.50% of the theoretical amount. If you desire to produce 700. grams of the C2H4O2 in a single batch, how many kg of C2H6O should you begin with?
Chemistry
1 answer:
Alexxandr [17]3 years ago
8 0
C2H6O + O2 ---> C2H4O2 + H2O

using the molar masses:-
24+ 6 + 16 g of C2H6O  produces 24 + 4 + 32 g C2H4O2    (theoretical)
 46 g produces 60g 

60 g C2H4O2 is produced from 46g C2H6O
1g      .     .................................46/60 g 
 700g     .................................    (46/60) * 700  Theoretically

But as the yield is only 7.5% 

the required amount is    ((46/60) * 700 ) / 0.075 =  7155.56 g 

=  7.156 kg to nearest gram.  Answer




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Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

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The intermediate balanced chemical reaction will be,

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(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=-41kJ

(3) CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206kJ

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\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3

\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)

\Delta H_{rxn}=-187.6kJ/mol

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

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\Delta H=\frac{q}{n}

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q=\Delta H\times n

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\Delta H = enthalpy change = -187.6 kJ/mol

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n = number of moles of coal = \frac{1.00\times 1000g}{12.00g/mol}=83.33mol

Now put all the given values in the above formula, we get:

q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ

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