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joja [24]
3 years ago
8

"The combustion of ethylene proceeds by the reaction C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g) When the rate of disappearance of O2 i

s 0.23 M s-1, the rate of disappearance of C2H4 is ________ M s-1."
Chemistry
1 answer:
Julli [10]3 years ago
7 0

Answer:

The rate of  disappearance of C_2H_4 is 0.0766 M/s.

Explanation:

C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)

Rate of the reaction = R

R=-\frac{1}{1}\frac{d[C_2H_4]}{dt}=-\frac{1}{3}\frac{d[O_2]}{dt}

[te]R=\frac{1}{2}\frac{d[CO_2]}{dt}=\frac{1}{2}\frac{d[H_2O]}{dt}[/tex]

Rate of disappearance of O_2 =0.23 M/s

Rate of disappearance of O_2=3\times R

0.23 M/s=3\times R

R =0.07666 M/s

Rate of disappearance of C_2H_4=1\times R

\frac{d[C_2H_4]}{dt}=1\times 0.23 m/s=0.0766 M/s

The rate of  disappearance of C_2H_4 is 0.0766 M/s.

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Answer: 41.5 mL

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Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

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V_s = volume of solution in L

Given : 59.4 g of H_2SO_4 in 100 g of solution  

moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

Volume of solution =\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml

Now put all the given values in the formula of molality, we get

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

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M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

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