Question

How much excess reactant is left over when 17.0g of potassium hydroxide

Answer 1

Approximately
`\(12.36 \, \text{grams}\)` of
`\(\text{KOH}\)` is left over as the excess reactant after the reaction.

To determine the excess reactant and the limiting reactant, we need to calculate the number of moles of each reactant using their molar masses and the balanced chemical equation.

The balanced chemical equation for the reaction between potassium hydroxide
`(\(\text{KOH}\))` and iron (III) nitrate
`\(\left(\text{Fe}(\text{NO}_3)_3\right)\)` is:

`\[\text{KOH} + \text{Fe}(\text{NO}_3)_3 \rightarrow \text{KNO}_3 + \text{Fe}(\text{OH})_3\]`

The molar masses are:

-
`\(\text{KOH}\) = \(56.11 \, \text{g/mol}\)`

-
`\(\text{Fe}(\text{NO}_3)_3\) = \(241.86 \, \text{g/mol}\)`

First, determine the number of moles for each reactant:

For
`\(\text{KOH}\)`:

`\[\text{Moles of KOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{17.0 \, \text{g}}{56.11 \, \text{g/mol}}\]`

For
`\(\text{Fe}(\text{NO}_3)_3\):`

`\[\text{Moles of Fe}(\text{NO}_3)_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0 \, \text{g}}{241.86 \, \text{g/mol}}\]`

Then, compare the mole ratios based on the balanced equation to identify the limiting reactant.

The balanced equation shows a 1:1 mole ratio between
`\(\text{KOH}\) and \(\text{Fe}(\text{NO}_3)_3\).`

- Moles of KOH:
`\(0.303 \, \text{mol}\)`

- Moles of Fe
`(\(\text{NO}_3)_3\): \(0.0828 \, \text{mol}\)`

Since the moles of
`\(\text{Fe}(\text{NO}_3)_3\)` are less than the moles of
`\(\text{KOH}\)` and both have a 1:1 ratio,
`\(\text{Fe}(\text{NO}_3)_3\)` is the limiting reactant.

To determine the excess amount of left over, us
`\(\text{KOH}\)`e the limiting reactant amount to find the actual amount of KOH used in the reaction.

Moles of
`\(\text{Fe}(\text{NO}_3)_3\)` used in the reaction
`\(= 0.0828 \, \text{mol}\)`

Since the mole ratio is 1:1, the moles of
`\(\text{KOH}\)` used are also
`\(0.0828 \, \text{mol}\).`

Therefore, the excess
`\(\text{KOH}\)` left over is:

`\[\text{Excess } \text{KOH} = \text{Total moles of KOH} - \text{Moles of KOH used}\]`

`\[\text{Excess } \text{KOH} = 0.303 \, \text{mol} - 0.0828 \, \text{mol}\]`

`\[\text{Excess } \text{KOH} = 0.2202 \, \text{mol}\]`

Finally, calculate the excess mass of
`\(\text{KOH}\)` left over using the molar mass:

`\[\text{Excess mass of KOH} = \text{Excess moles} * \text{Molar mass of KOH}\]`

`\[\text{Excess mass of KOH} = 0.2202 \, \text{mol} * 56.11 \, \text{g/mol}\]`

`\[\text{Excess mass of KOH} \approx 12.36 \, \text{g}\]`

Question:

How much excess reactant is left over when
`$17.0 \mathrm{~g}$` of potassium hydroxide
`$(\mathrm{KOH})$` reacts with
`$20.0 . \mathrm{g}$` of iron (III) nitrate
`$\left(\mathrm{Fe}\left(\mathrm{NO}_3\right)_3\right)$`?