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shtirl [24]
3 years ago
11

O2 + 2C0-2CO2

Chemistry
1 answer:
andreev551 [17]3 years ago
5 0

Answer: 1. CO is the limiting reagent and O_2 is the excess reagent.

2. 2.91 g of excess reagent

3. 4.66 g of carbon dioxide produced.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} CO=\frac{2.98g}{28g/mol}=0.106moles

\text{Moles of} O_2=\frac{4.62g}{32g/mol}=0.144moles

O_2+2CO\rightarrow 2CO_2  

According to stoichiometry :

2 moles of CO require = 1 mole of O_2

Thus 0.106 moles of CO will require=\frac{1}{2}\times 0.106=0.053moles  of O_2

Thus CO is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Amount of excess reagent = ( 0.144-0.053) moles = 0.091 moles

Mass of excess reagent O_2=moles\times {\text {Molar mass}}=0.091moles\times 32g/mol=2.91g

As 2 moles of CO give = 2 moles of CO_2

Thus 0.106 moles of CO give =\frac{2}{2}\times 0.106=0.106moles  of CO_2

Mass of CO_2=moles\times {\text {Molar mass}}=0.106moles\times 44g/mol=4.66g

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a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

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Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

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ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

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ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

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ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

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