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2 years ago

2. Identify the limiting reactant when 4.68 g of iron reacts with 2.88 g of sulfur to produce Fes.

Chemistry

1 answer:

Vinvika [58]2 years ago

6 0

Answer:

Iron is limiting reactant

Explanation:

Based on the reaction:

Fe + S → FeS

1 mole of iron reacts per mole of Sulfur

To solve this question we must convert the mass of each reactant to moles using molar masses of each reactant. As the reaction is 1:1, the reactant with the lower amount of moles is limiting reactant.

Moles Fe -Molar mass: 55.845g/mol-

4.68g * (1mol / 55.845g) = 0.0838 moles

Moles S -Molar mass: 32.065g/mol-

2.88g * (1mol / 32.065g) = 0.0898 moles

As the amount of moles of Fe < Moles S,

Iron is limiting reactant

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In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

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Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .

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