Problem 4: 3CaCO_(3)+2FePO_(4)arrow Ca_(3)(PO_(4))_(2)+Fe_(2)(CO_(3))_(3) Given: 50 grams of 1 point calcium carbonate and 15 grams of Iron (III) phosphate. What is the Maximum amount of Fe_(2)(CO_(3))_(3) can be made? __ A convof your responses will be emailed to 4806987222( students.ocps.net.

To solve this problem, we need to determine the limiting reactant and then calculate the maximum amount of iron(III) carbonate (\(Fe_2(CO_3)_3\)) that can be produced.<br /><br />First, let's calculate the molar masses of the reactants and products:<br /><br />1. **Calcium Carbonate (\(CaCO_3\))**:<br /> - \(Ca: 40.08 \, \text{g/mol}\)<br /> - \(C: 12.01 \, \text{g/mol}\)<br /> - \(O_3: 3 \times 16.00 \, \text{g/mol} = 48.00 \, \text{g/mol}\)<br /> - Total: \(40.08 + 12.01 + 48.00 = 100.09 \, \text{g/mol}\)<br /><br />2. **Iron(III) Phosphate (\(FePO_4\))**:<br /> - \(Fe: 55.85 \, \text{g/mol}\)<br /> - \(P: 30.97 \, \text{g/mol}\)<br /> - \(O_4: 4 \times 16.00 \, \text{g/mol} = 64.00 \, \text{g/mol}\)<br /> - Total: \(55.85 + 30.97 + 64.00 = 150.82 \, \text{g/mol}\)<br /><br />3. **Iron(III) Carbonate (\(Fe_2(CO_3)_3\))**:<br /> - \(Fe_2: 2 \times 55.85 \, \text{g/mol} = 111.70 \, \text{g/mol}\)<br /> - \(C_3: 3 \times 12.01 \, \text{g/mol} = 36.03 \, \text{g/mol}\)<br /> - \(O_9: 9 \times 16.00 \, \text{g/mol} = 144.00 \, \text{g/mol}\)<br /> - Total: \(111.70 + 36.03 + 144.00 = 291.73 \, \text{g/mol}\)<br /><br />Next, we convert the masses of the reactants to moles:<br /><br />1. **Calcium Carbonate**:<br /> \[<br /> \text{Moles of } CaCO_3 = \frac{50 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.500 \, \text{mol}<br /> \]<br /><br />2. **Iron(III) Phosphate**:<br /> \[<br /> \text{Moles of } FePO_4 = \frac{15 \, \text{g}}{150.82 \, \text{g/mol}} \approx 0.099 \, \text{mol}<br /> \]<br /><br />Now, we use the stoichiometry of the balanced equation to determine the limiting reactant:<br /><br />\[<br />3CaCO_3 + 2FePO_4 \rightarrow Ca_3(PO_4)_2 + Fe_2(CO_3)_3<br />\]<br /><br />From the equation, 3 moles of \(CaCO_3\) react with 2 moles of \(FePO_4\). Therefore, the ratio of \(CaCO_3\) to \(FePO_4\) should be 3:2.<br /><br />We need to check which reactant is the limiting reactant by comparing the actual mole ratio to the stoichiometric ratio:<br /><br />\[<br />\text{Actual ratio} = \frac{0.500 \, \text{mol}}{0.099 \, \text{mol}} \approx 5.05<br />\]<br /><br />\[<br />\text{Stoichiometric ratio} = \frac{3}{2} = 1.5<br />\]<br /><br />Since the actual ratio (5.05) is greater than the stoichiometric ratio (1.5), \(FePO_4\) is the limiting reactant.<br /><br />Now, we calculate the maximum amount of \(Fe_2(CO_3)_3\) that can be produced from the limiting reactant:<br /><br />\[<br />\text{Moles of } Fe_2(CO_3)_3 = \frac{0.099 \, \text{mol} \times 1 \, \text{mol} \, Fe_2(CO_3)_3}{2 \, \text{mol} \, FePO_4} \approx 0.0495 \, \text{mol}<br />\]<br /><br />Finally, we convert the moles of \(Fe_2(CO_3)_3\) to grams:<br /><br />\[<br />\text{Mass of } Fe_2(CO_3)_3 = 0.0495 \, \text{mol} \times