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Leya [2.2K]
3 years ago
6

Problem PageQuestion Gaseous ethane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If of w

ater is produced from the reaction of of ethane and of oxygen gas, calculate the percent yield of water.
Chemistry
1 answer:
Nata [24]3 years ago
5 0

Answer:

<h2> = ( 1.08 / 2.2 ) 100% = 49%</h2>

Explanation:

Balanced Equation: 2CH₃CH₃(g) + 5O₂(g) → 2CO₂(g) + 6H₂O(g)

Calculate moles of CH₃CH₃ and O₂

1.2 ₃₃ ( 1 ₃₃/ 30.0694 ₃₃ ) = 0.040 ₃₃

8.6 ₂ ( 1 2/ 31.998 ₂ ) = 0.27 ₃₃

Find limiting reagent  0.040 ₃₃ ( 5 ₂/ 2 ₃₃ ) = 0.10 ₂

CH₃CH₃ is the limiting Reagent

CH₃CH₃ (L.R.) O₂ CO₂ H₂O

Initial (mol) 0.040 0.27 0 0

Change (mol) -2x=-0 -5x= -0.10  +2x=+0.040 +6x=+0.12

Final (mol) 0 0.117 0.040 0.12

0.040 − 2 = 0 = 0.020

Determine percent yield

0.12 ₂ ( 18.0148 ₂ /1 ₂ ) = 2.2 ₂  

= ( 1.08 / 2.2 ) 100% = 49%

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(a)  As the given chemical reaction equation is as follows.

           OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}

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              Rate = K \times [OCl^{-}] \times [l^{-}]

(b)  Since, the rate equation is as follows.

                    Rate = K [OCl^{-}][l^{-}]

Let us assume that ([OCl^{-}] = [l^{-}])

Putting the given values into the above equation as follows.

             1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2

            1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})

                   K = \frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}

                      = 60.4 M^{-1}sec^{-1}

Hence, the value of rate constant for the given reaction is 60.4 M^{-1}sec^{-1} .

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                Rate = K [OCl^{-}][l^{-}]

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                        = 6.52 \times 10^{5}

Therefore, rate when [OCl^{-}] = 1.8 \times 10^{3} M and [I^{-}]= 6.0 \times 10^{4} M is  6.52 \times 10^{5}.

8 0
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