Problem 3: 3CaCO_(3)+2FePO_(4)arrow Ca_(3)(PO_(4))_(2)+Fe_(2)(CO_(3))_(3) Given: 50.0 grams of calcium carbonate and 15.0 grams of Iron (III) phosphate. What is the Maximum amount of Fe_(2)(CO_(3))_(3) can be made? 0.0497molFe_(2)(CO_(3))_(3) 1.12LFe_(2)(CO_(3))_(3) 14.6 q Fe_(2)(CO_(3))_(3) 3.01E22 molecules Fe_(2)(CO_(3))_(3) 0.166 mol Fe_(2)(CO_(3))_(3) 3.73 L Fe_(2)(CO_(3))_(3) 1.00E23 molecules Fe_(2)(CO_(3))_(3)

To solve this problem, we need to determine the limiting reactant and then calculate the maximum amount of $Fe_{2}(CO_{3})_{3}$ that can be produced.<br /><br />Given information:<br />- 50.0 grams of calcium carbonate ($CaCO_{3}$)<br />- 15.0 grams of Iron (III) phosphate ($FePO_{4}$)<br /><br />Step 1: Calculate the molar mass of each reactant.<br />- Molar mass of $CaCO_{3}$ = 100.09 g/mol<br />- Molar mass of $FePO_{4}$ = 150.82 g/mol<br /><br />Step 2: Calculate the number of moles of each reactant.<br />- Moles of $CaCO_{3}$ = 50.0 g / 100.09 g/mol = 0.500 mol<br />- Moles of $FePO_{4}$ = 15.0 g / 150.82 g/mol = 0.099 mol<br /><br />Step 3: Determine the limiting reactant.<br />- The balanced chemical equation is:<br /> $3CaCO_{3} + 2FePO_{4} \rightarrow Ca_{3}(PO_{4})_{2} + Fe_{2}(CO_{3})_{3}$<br />- The stoichiometric ratio of $CaCO_{3}$ to $FePO_{4}$ is 3:2.<br />- The actual ratio of $CaCO_{3}$ to $FePO_{4}$ is 0.500 mol / 0.099 mol = 5.05.<br />- Since the actual ratio is greater than the stoichiometric ratio, $FePO_{4}$ is the limiting reactant.<br /><br />Step 4: Calculate the maximum amount of $Fe_{2}(CO_{3})_{3}$ that can be produced.<br />- The balanced chemical equation shows that 2 moles of $FePO_{4}$ produce 1 mole of $Fe_{2}(CO_{3})_{3}$.<br />- Therefore, the maximum amount of $Fe_{2}(CO_{3})_{3}$ that can be produced is 0.099 mol / 2 = 0.0495 mol.<br /><br />Therefore, the maximum amount of $Fe_{2}(CO_{3})_{3}$ that can be made is 0.0495 mol.