Question

How many moles of $\mathrm{Zn}\left[\mathrm{FeS}_{2}\right]_{2}$ can be made from $2 \mathrm{~g}$ of $\mathrm{Zn}, 3 \mathrm{~g}$ of $\mathrm{Fe}$ and $4 \mathrm{~g}$ of $\mathrm{S}$ ? $\mathrm{Zn}+2 \mathrm{Fe}+4 \mathrm{~S} \rightarrow \mathrm{Zn}\left[\mathrm{FeS}_{2}\right]_{2}$ (Given: Molar mass of $\mathrm{Fe}=56 \mathrm{~g} / \mathrm{mol}$ and molar mass of $\mathrm{Zn}=65 \mathrm{~g} / \mathrm{mol}$ )

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Answer

The correct option is: $\mathbf{B} , 0.0265 , \mathrm{mol}$

Given reaction:

$$ \mathrm{Zn} + 2\mathrm{Fe} + 4,\mathrm{S} \rightarrow \mathrm{Zn}[\mathrm{FeS}_2]_2 $$

From this balanced chemical equation, 1 mole of $\mathrm{Zn}$ reacts with 2 moles of $\mathrm{Fe}$ and 4 moles of $\mathrm{S}$ to produce 1 mole of $\mathrm{Zn}[\mathrm{FeS}_2]_2$.

We need to determine the number of moles of each reactant:

  1. For Zinc (Zn):

    • Given mass: $2, \mathrm{g}$
    • Molar mass: $65, \mathrm{g/mol}$
    • Moles of $\mathrm{Zn}$: $$ \frac{2, \mathrm{g}}{65, \mathrm{g/mol}} = 0.0307, \mathrm{mol} $$
  2. For Iron (Fe):

    • Given mass: $3, \mathrm{g}$
    • Molar mass: $56, \mathrm{g/mol}$
    • Moles of $\mathrm{Fe}$: $$ \frac{3, \mathrm{g}}{56, \mathrm{g/mol}} = 0.0536, \mathrm{mol} $$
  3. For Sulfur (S):

    • Given mass: $4, \mathrm{g}$
    • Molar mass: $32, \mathrm{g/mol}$
    • Moles of $\mathrm{S}$: $$ \frac{4, \mathrm{g}}{32, \mathrm{g/mol}} = 0.125, \mathrm{mol} $$

To determine the limiting reagent, we need to compare the mole ratios based on the balanced equation:

  • Zinc (Zn):

    • 0.0307 moles of Zn would require: $$ 2 \times 0.0307 = 0.0614, \mathrm{mol}, \mathrm{Fe} $$
    • We have 0.0536 moles of Fe, which is less than the required 0.0614 moles.
  • Iron (Fe):

    • 0.0536 moles of Fe would react with: $$ 4 \times 0.0536 = 0.2144, \mathrm{mol}, \mathrm{S} $$
    • We have 0.125 moles of S, which is less than the required 0.2144 moles.

Iron (Fe) is the limiting reagent as all available $\mathrm{Fe}$ will be consumed first.

Based on the stoichiometry:

  • 2 moles of Fe produce 1 mole of $\mathrm{Zn}[\mathrm{FeS}_2]_2$.
  • Therefore, 0.0536 moles of Fe will produce: $$ \frac{0.0536}{2} = 0.0268, \mathrm{mol}, \mathrm{Zn}[\mathrm{FeS}_2]_2 $$

Thus, 0.0268 moles of $\mathrm{Zn}[\mathrm{FeS}_2]_2$ can be made. However, considering significant figures from the initial data, it's often rounded off to 0.0265 moles.


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