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Calculate N (number of moles) using the provided parameters and formula: N = m/RAM = (I t) / (z F).

Chemistry

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Answer

0.0166 mol\text{0.0166 mol}

Step-by-step solution

6 steps

Here's the solution using the information provided in the image. The question asks to find NN (number of moles). The image provides two ways to calculate NN.

Given Parameters from the image:

  • Relative Atomic Mass (RAM\text{RAM}) = 119 g/mol
  • Mass (mm) = 26.6 g
  • Current (II) = 8 A
  • Time (tt) = 400 s
  • Valency (zz) = 2
  • Faraday's constant (FF) = 96500 C/mol

We will calculate NN using both parts of the given formula: N=mRAM=I×tz×FN = \frac{m}{\text{RAM}} = \frac{I \times t}{z \times F}.

Method 1: Calculate NN using mass and relative atomic mass.

N=mRAMN = \frac{m}{\text{RAM}}
  1. Substitute the given values.

    N=26.6 g119 g/molN = \frac{26.6 \text{ g}}{119 \text{ g/mol}}
  2. Calculate the value of NN.

    N≈0.223529 molN \approx 0.223529 \text{ mol}

    Method 2: Calculate NN using current, time, valency, and Faraday's constant.

    N=I×tz×FN = \frac{I \times t}{z \times F}
  3. Substitute the given values.

    N=8 A×400 s2×96500 C/molN = \frac{8 \text{ A} \times 400 \text{ s}}{2 \times 96500 \text{ C/mol}}
  4. Perform the multiplication in the numerator.

    N=3200 C2×96500 C/molN = \frac{3200 \text{ C}}{2 \times 96500 \text{ C/mol}}
  5. Perform the multiplication in the denominator.

    N=3200 C193000 C/molN = \frac{3200 \text{ C}}{193000 \text{ C/mol}}
  6. Calculate the value of NN.

    N≈0.016580 molN \approx 0.016580 \text{ mol}

    There is a significant difference between the values of NN calculated by the two methods. This indicates that the given parameters (mm, II, tt) are not consistent with each other for the same electrochemical process.

    If we assume the question intends for us to calculate NN using the current and time, as it was the focus of the previous interaction:

    The number of moles (NN) calculated from current and time is approximately 0.0166 mol\boxed{\text{0.0166 mol}}.

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