underline ( )Na_(2)(CO_(3))+underline ( )Fe(OH)_(3)arrow underline ( )Na(OH)+underline ( )Fe_(2)(CO_(3))_(3) 4. How many moles of Na_(2)(CO_(3)) are needed to react with 25.5 g of Fe(OH)_(3)

To determine the number of moles of \( Na_2(CO_3) \) needed to react with 25.5 g of \( Fe(OH)_3 \), we need to follow these steps:<br /><br />1. **Write the balanced chemical equation:**<br /> \[<br /> 3Na_2(CO_3) + 2Fe(OH)_3 \rightarrow 6NaOH + Fe_2(CO_3)_3<br /> \]<br /><br />2. **Calculate the molar mass of \( Fe(OH)_3 \):**<br /> - Molar mass of Fe: \( 55.85 \, \text{g/mol} \)<br /> - Molar mass of O: \( 16.00 \, \text{g/mol} \)<br /> - Molar mass of H: \( 1.01 \, \text{g/mol} \)<br /><br /> \[<br /> \text{Molar mass of } Fe(OH)_3 = 55.85 + 3(16.00 + 1.01) = 106.88 \, \text{g/mol}<br /> \]<br /><br />3. **Convert the mass of \( Fe(OH)_3 \) to moles:**<br /> \[<br /> \text{Moles of } Fe(OH)_3 = \frac{25.5 \, \text{g}}{106.88 \, \text{g/mol}} \approx 0.238 \, \text{moles}<br /> \]<br /><br />4. **Use the stoichiometry of the balanced equation to find the moles of \( Na_2(CO_3) \):**<br /> According to the balanced equation, 2 moles of \( Fe(OH)_3 \) react with 3 moles of \( Na_2(CO_3) \).<br /><br /> \[<br /> \text{Moles of } Na_2(CO_3) = 0.238 \, \text{moles} \times \frac{3 \, \text{moles of } Na_2(CO_3)}{2 \, \text{moles of } Fe(OH)_3} \approx 0.357 \, \text{moles}<br /> \]<br /><br />Therefore, approximately 0.357 moles of \( Na_2(CO_3) \) are needed to react with 25.5 g of \( Fe(OH)_3 \).