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Ratling [72]
2 years ago
7

How many Faradays of electricity are carried by 1.505×10^23 electronsk​

Chemistry
1 answer:
aksik [14]2 years ago
3 0

Mole of electron required by 1.505 *10^{23} mole is 2.5* 10  ^{-1}

  • Faraday law expressed how the  change that is been being produced by a current at an electrode-electrolyte interface is related and  proportional to the quantity of electricity  that is been used.

  • There is one mole of electron required for 1 Faraday of electricity.

  • Avogadro constant is 6.02*10^{23}

  • Mole of electron can be calculated by dividing the number of electron by avogadro's constant.

=\frac{1.505*10^{23} }{6.02*10^{23} }

= 2.5* 10^{-1}  Faraday  of electricity

Therefore, it requires  2.5*10^{-1} Faraday of electricity for the 1.505 *  10^{23}  mole.

Learn more at: brainly.com/question/1640558?referrer=searchResults

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<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol

For the given chemical reaction:

2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = \frac{12}{2}\times 0.778=4.668 mol of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

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