Question

Solid iron (III) hydroxide decomposes to produce iron (III) oxide and water vapor. Write a balanced equation. If 0.75 L of water vapor are produced at STP,

Fe(OH)3 --> Fe2O3 + H2O (YOU MUST BALANCE THIS BEFORE STARTING)

a. How many grams of iron (III) hydroxide were used? _________ grams of iron (III) hydroxide (3 sig figs)


b. How many grams of iron (III) oxide were produced? __________ grams of iron (III) oxide (3 sig figs)


50 points

Answer 1

Balanced equation: 2Fe(OH)₃ → Fe₂O₃ + 3H₂O

a) 2.39 g

b) 1.78 g

Step-by-step explanation:

Given unbalanced chemical equation:

Fe(OH)₃ → Fe₂O₃ + H₂O

To balance a chemical equation, we need to make sure that there are the same number of atoms of each element on both sides of the equation. We can do this by adding coefficients in front of a chemical symbol or formula where needed.

Therefore, the balanced equation is:

`\large\boxed{\sf 2Fe(OH)_3 \rightarrow Fe_2O_3 + 3H_2O}`

`\hrulefill`

Part (a)

Given that 0.75 L of water vapor (H₂O) is produced at STP (Standard Temperature and Pressure), which is 273.15 K and 1 atm, we can calculate the number of moles of water vapor produced using the ideal gas law:

`\large\boxed{PV = nRT}`

where:

• P = 1 atm (pressure)
• V = 0.75 L (volume)
• n is the number of moles of H₂O we want to calculate.
• R is the ideal gas constant (0.082 L·atm/mol·K)
• T = 273.15 K (temperature at STP).

Solving for n:

`\begin{aligned}n &= (PV)/(RT)\\\\n&= (1 \cdot 0.75)/(0.082\cdot 273.15)\\\\n&\approx 0.0334846841 \;\rm mol\end{aligned}`

Therefore, the number of moles of water vapor produced is 0.0335 moles (3 s.f.).

Using the stoichiometric coefficients from the balanced equation, we can relate the moles of water vapor produced to the moles of iron (III) hydroxide (Fe(OH)₃) used.

From the balanced equation, 2 moles of Fe(OH)₃ produce 3 moles of H₂O. Therefore, the number of moles of Fe(OH)₃ is:

`\phantom{w.}\rm 0.0334846841\; moles\;of\;H_2O * \frac{2 \text{ moles of } Fe(OH)_3}{3 \text{ moles of } H_2O}\\\\ \approx 0.0223231227 \text{ moles of } Fe(OH)_3`

Now, calculate the molar mass of Fe(OH)₃:

`\phantom{w.}\rm Molar\; mass\; of\; Fe(OH)_3\\\\=\rm (1 * Atomic\; mass \;of\; Fe) + (3 * Atomic\; mass\; of\; O) + (3 * Atomic \;mass\; of \;H)\\\\\rm \approx (1 * 55.845\;g/mol) + (3 * 15.999\; g/mol) + (3 * 1.008\;g/mol)\\\\\rm \approx 106.866\; g/mol`

Finally, calculate the mass of Fe(OH)₃ by multiplying the number of moles by the molar mass:

`\begin{aligned}\rm Mass\; of\; Fe(OH)_3 &= \rm 0.0223231227 \; mol * 106.866\; g/mol\\&= \rm 2.38558283\; g\\&\approx \rm 2.39\;g\;(3\;s.f.)\end{aligned}`

Therefore, approximately 2.39 grams of iron (III) hydroxide were used.

`\hrulefill`

Part (b)

From the balanced equation, 2 moles of Fe(OH)₃ produce 1 mole of Fe₂O₃. Therefore, the moles of Fe₂O₃ produced is:

`\phantom{w.}0.0223231227 \text{ moles of } \rm Fe(OH)_3 * \frac{1 \text{ mole of } Fe_2O_3}{2 \text{ moles of } Fe(OH)_3}\\\\ \approx 0.0111615614\text{ moles of } Fe_2O_3`

Now, calculate the molar mass of Fe₂O₃:

`\phantom{w.}\rm Molar\; mass\; of\; Fe_2O_3\\\\=\rm (2 * Atomic\; mass \;of\; Fe) + (3 * Atomic\; mass\; of\; O) \\\\\rm \approx (2 * 55.845\;g/mol) + (3 * 15.999\; g/mol) \\\\\rm \approx 159.687\; g/mol`

Finally, calculate the mass of Fe₂O₃ by multiplying the number of moles by the molar mass:

`\begin{aligned}\rm Mass\; of\; Fe_2O_3&= \rm 0.0111615614 \; mol * 159.687\; g/mol\\&= \rm 1.78235625\; g\\&\approx \rm 1.78\;g\;(3\;s.f.)\end{aligned}`

Therefore, approximately 1.78 grams of iron (III) oxide were used.