To determine how many grams of iron(II) carbonate ([tex]\(FeCO_3\)[/tex]) can be produced from the given reaction, we will follow these steps:
1. Identify the Given Data:
- Volume of [tex]\(FeCl_2\)[/tex] solution: 1.24 liters
- Molarity of [tex]\(FeCl_2\)[/tex] solution: 2.00 M
- Molar mass of [tex]\(FeCO_3\)[/tex]: 115.854 g/mol
2. Calculate the Moles of [tex]\(FeCl_2\)[/tex] in Solution:
Using the formula:
[tex]\[
\text{moles of } FeCl_2 = \text{Volume (L)} \times \text{Molarity (M)}
\][/tex]
[tex]\[
\text{moles of } FeCl_2 = 1.24 \text{ L} \times 2.00 \text{ M} = 2.48 \text{ moles}
\][/tex]
3. Determine the Moles of [tex]\(FeCO_3\)[/tex] Produced:
The balanced chemical equation is:
[tex]\[
FeCl_2( aq ) + Na_2 CO_3(s) \rightarrow FeCO_3(s) + 2NaCl(aq)
\][/tex]
According to the stoichiometry of the reaction, 1 mole of [tex]\(FeCl_2\)[/tex] reacts to produce 1 mole of [tex]\(FeCO_3\)[/tex]. Hence, the moles of [tex]\(FeCl_2\)[/tex] are equal to the moles of [tex]\(FeCO_3\)[/tex] produced:
[tex]\[
\text{moles of } FeCO_3 = 2.48 \text{ moles}
\][/tex]
4. Calculate the Mass of [tex]\(FeCO_3\)[/tex] Produced:
Using the molar mass of [tex]\(FeCO_3\)[/tex]:
[tex]\[
\text{mass of } FeCO_3 = \text{moles of } FeCO_3 \times \text{molar mass of } FeCO_3
\][/tex]
[tex]\[
\text{mass of } FeCO_3 = 2.48 \text{ moles} \times 115.854 \text{ g/mol}
\][/tex]
[tex]\[
\text{mass of } FeCO_3 = 287.31792 \text{ grams}
\][/tex]
5. Express the Answer to Three Significant Figures:
The mass of iron(II) carbonate that can be produced is:
[tex]\[
287 \text{ grams}
\][/tex]
Therefore, the reaction can produce [tex]\( \boxed{287} \)[/tex] grams of iron(II) carbonate.
