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?
Balance the equation
Balance the equation
Answers
GPT-4o mini
To start, we need to write and balance the equation for the reaction between aqueous iron(II) hydroxide \((\text{Fe(OH)}_2)\) and aqueous copper(I) carbonate \((\text{Cu}_2\text{CO}_3)\) to produce aqueous iron(II) carbonate \((\text{FeCO}_3)\) and solid copper(I) hydroxide \((\text{CuOH})\).
The unbalanced equation is:
\[
\text{Fe(OH)}_2(aq) + \text{Cu}_2\text{CO}_3(aq) \rightarrow \text{FeCO}_3(aq) + \text{CuOH}(s)
\]
### Balancing the equation:
1. On the left side, we have 1 Fe and 1 CO₃ from \( \text{Cu}_2\text{CO}_3\) (which contributes 2 Cu).
2. On the right side, we have 1 Fe connected to 1 CO₃ and 2 Cu in hydroxide form.
Thus, we can balance it as follows:
\[
\text{Fe(OH)}_2(aq) + \text{Cu}_2\text{CO}_3(aq) \rightarrow \text{FeCO}_3(aq) + 2\text{CuOH}(s)
\]
### The balanced equation is:
\[
\text{Fe(OH)}_2(aq) + \text{Cu}_2\text{CO}_3(aq) \rightarrow \text{FeCO}_3(aq) + 2\text{CuOH}(s)
\]
### Moles of Iron(II) Hydroxide:
Next, we need to find out how many moles of iron(II) hydroxide are in 50.0 g.
1. The molar mass of \(\text{Fe(OH)}_2\):
- Iron (Fe): \(55.85 \, \text{g/mol}\)
- Oxygen (O): \(16.00 \, \text{g/mol} \times 2 = 32.00 \, \text{g/mol}\)
- Hydrogen (H): \(1.01 \, \text{g/mol} \times 2 = 2.02 \, \text{g/mol}\)
- Total for \(\text{Fe(OH)}_2\):
\[ 55.85 + 32.00 + 2.02 = 89.87 \, \text{g/mol} \]
2. Calculate the moles of \(\text{Fe(OH)}_2\):
\[
\text{moles of Fe(OH)}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{50.0 \, \text{g}}{89.87 \, \text{g/mol}} \approx 0.556 \, \text{mol}
\]
### Moles of Iron(II) Carbonate Produced:
From the balanced equation, we can see that the mole ratio of iron(II) hydroxide to iron(II) carbonate is 1:1. This means that the number of moles of iron(II) carbonate produced will be equal to the number of moles of iron(II) hydroxide reacted.
Thus, the moles of iron(II) carbonate produced would also be \(0.556 \, \text{mol}\).
### Conclusion:
If 50.0 g of iron(II) hydroxide reacts completely, approximately \(0.556\) moles of iron(II) carbonate will be produced.
The unbalanced equation is:
\[
\text{Fe(OH)}_2(aq) + \text{Cu}_2\text{CO}_3(aq) \rightarrow \text{FeCO}_3(aq) + \text{CuOH}(s)
\]
### Balancing the equation:
1. On the left side, we have 1 Fe and 1 CO₃ from \( \text{Cu}_2\text{CO}_3\) (which contributes 2 Cu).
2. On the right side, we have 1 Fe connected to 1 CO₃ and 2 Cu in hydroxide form.
Thus, we can balance it as follows:
\[
\text{Fe(OH)}_2(aq) + \text{Cu}_2\text{CO}_3(aq) \rightarrow \text{FeCO}_3(aq) + 2\text{CuOH}(s)
\]
### The balanced equation is:
\[
\text{Fe(OH)}_2(aq) + \text{Cu}_2\text{CO}_3(aq) \rightarrow \text{FeCO}_3(aq) + 2\text{CuOH}(s)
\]
### Moles of Iron(II) Hydroxide:
Next, we need to find out how many moles of iron(II) hydroxide are in 50.0 g.
1. The molar mass of \(\text{Fe(OH)}_2\):
- Iron (Fe): \(55.85 \, \text{g/mol}\)
- Oxygen (O): \(16.00 \, \text{g/mol} \times 2 = 32.00 \, \text{g/mol}\)
- Hydrogen (H): \(1.01 \, \text{g/mol} \times 2 = 2.02 \, \text{g/mol}\)
- Total for \(\text{Fe(OH)}_2\):
\[ 55.85 + 32.00 + 2.02 = 89.87 \, \text{g/mol} \]
2. Calculate the moles of \(\text{Fe(OH)}_2\):
\[
\text{moles of Fe(OH)}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{50.0 \, \text{g}}{89.87 \, \text{g/mol}} \approx 0.556 \, \text{mol}
\]
### Moles of Iron(II) Carbonate Produced:
From the balanced equation, we can see that the mole ratio of iron(II) hydroxide to iron(II) carbonate is 1:1. This means that the number of moles of iron(II) carbonate produced will be equal to the number of moles of iron(II) hydroxide reacted.
Thus, the moles of iron(II) carbonate produced would also be \(0.556 \, \text{mol}\).
### Conclusion:
If 50.0 g of iron(II) hydroxide reacts completely, approximately \(0.556\) moles of iron(II) carbonate will be produced.