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FromTheMoon [43]
1 year ago
15

Balance the following skeleton reactions and identify the oxidizing and reducing agents:(b) Mn²⁺(aq) + BiO₃⁻(aq) → MnO₄⁻(aq) + B

i³⁺(aq) [acidic]
Chemistry
1 answer:
amm18121 year ago
5 0

The balanced skeleton reactions are given below

2Mn²⁺(aq) +5BiO₃⁻(aq) +  14H^{+}(aq)  →  2MnO₄⁻(aq) + 5Bi³⁺(aq) + 7H_{2}O(l)

The oxidising agent and  reducing agents are Bi and Mn.

There are following steps involve in balance redox reaction in acidic medium .

  • First divide the reaction into two half reactions .
  • Balance the elements other than oxygen and hydrogen atom.
  • Balance the oxygen atom by adding H_{2}O on that side where number of oxygen  atom is less .
  • Balance the hydrogen atom by adding H^{+} on another side of in which water attached .
  • Balance the charge by adding e^{-} .
  • Add the two half reaction  .

The species which reduces the other species in the redox reaction is called oxidizing agent . Similarly , the species which oxidize the other species in the redox reaction  is called reducing agent .

To learn more about reducing agents please click here ,

brainly.com/question/2890416?

#SPJ4

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Answer :

(a) The moles of water produced are 145.35 moles.

(b) The mass of oxygen needed are 3080.8 grams.

<u>Solution for part (a) : Given,</u>

Moles of C_8H_{18} = 16.15 moles

First we have to calculate the moles of H_2O

The balanced chemical reaction is,

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

From the balanced reaction we conclude that

As, 2 moles of C_8H_{18} react to give 18 moles of H_2O

So, 16.15 moles of C_8H_{18} react to give \frac{16.15}{2}\times 18=145.35 moles of H_2O

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<u>Solution for part (b) : Given,</u>

Mass of C_8H_{18} = 878 g

Molar mass of C_8H_{18} = 114 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_8H_{18}.

\text{ Moles of }C_8H_{18}=\frac{\text{ Mass of }C_8H_{18}}{\text{ Molar mass of }C_8H_{18}}=\frac{878g}{114g/mole}=7.702moles

Now we have to calculate the moles of O_2

The balanced chemical reaction is,

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

From the balanced reaction we conclude that

As, 2 moles of C_8H_{18} react with 25 moles of O_2

So, 7.702 moles of C_8H_{18} react with \frac{7.702}{2}\times 25=96.275 moles of O_2

Now we have to calculate the mass of O_2.

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(96.275moles)\times (32g/mole)=3080.8g

The mass of oxygen needed are 3080.8 grams.

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