Question

Aqueous iron(II) hydroxide reacts with aqueous copper(I) carbonate to form aqueous iron(II) carbonate and solid copper(I) hydroxide. If 50.0g of iron(II) hydroxide were fully reacted, how many moles of iron(II) carbonate would be produced?

Answer 1

To determine how many moles of iron(II) carbonate can be produced from the complete reaction of 50.0 g of iron(II) hydroxide, we first need to consider the balanced chemical reaction between iron(II) hydroxide and copper(I) carbonate. The reaction can be represented as follows:

\[
\text{Fe(OH)}_2 (aq) + \text{Cu}_2\text{CO}_3 (aq) \rightarrow \text{FeCO}_3 (aq) + \text{CuOH} (s)
\]

This shows that one mole of iron(II) hydroxide reacts with one mole of copper(I) carbonate to produce one mole of iron(II) carbonate and one mole of solid copper(I) hydroxide.

Next, let's calculate the number of moles of iron(II) hydroxide (\(\text{Fe(OH)}_2\)) in 50.0 g. The molar mass of iron(II) hydroxide can be calculated as follows:

- Atomic mass of Fe: 55.85 g/mol
- Atomic mass of O: 16.00 g/mol
- Atomic mass of H: 1.01 g/mol

So, the molar mass of \(\text{Fe(OH)}_2\) is:

\[
\text{Molar mass of Fe(OH)}_2 = 55.85 + 2(16.00 + 1.01) = 55.85 + 2(17.01) = 55.85 + 34.02 = 89.87 \, \text{g/mol}
\]

Now, we can calculate the number of moles of iron(II) hydroxide in 50.0 g:

\[
\text{Number of moles of Fe(OH)}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{50.0 \, \text{g}}{89.87 \, \text{g/mol}} \approx 0.555 \, \text{moles}
\]

Since the balanced reaction indicates that 1 mole of iron(II) hydroxide produces 1 mole of iron(II) carbonate, the number of moles of iron(II) carbonate produced will also be approximately 0.555 moles.

Thus, the final answer is:

\[
\text{Moles of iron(II) carbonate produced} \approx 0.555 \, \text{moles}
\]