Problem 4: 3CaCO_(3)+2FePO_(4)arrow Ca_(3)(PO_(4))_(2)+Fe_(2)(CO_(3))_(3) Given: 50 grams of 1 calcium carbonate and 15 grams of Iron (III) phosphate. Find Which of the reagents is the limiting reagent? calcium carbonate iron III phosphate calcium phosphate iron III carbonate

To determine the limiting reagent, we need to calculate the number of moles of each reactant and then compare them to see which one is completely consumed in the reaction.<br /><br />Given:<br />- 50 grams of calcium carbonate (CaCO3)<br />- 15 grams of Iron (III) phosphate (FePO4)<br /><br />First, let's calculate the molar mass of each compound:<br /><br />1. Calcium carbonate (CaCO3):<br /> - Ca: 40.08 g/mol<br /> - C: 12.01 g/mol<br /> - O: 16.00 g/mol (3 atoms)<br /> - Molar mass of CaCO3 = 40.08 + 12.01 + (16.00 * 3) = 100.09 g/mol<br /><br />2. Iron (III) phosphate (FePO4):<br /> - Fe: 55.85 g/mol<br /> - P: 30.97 g/mol<br /> - O: 16.00 g/mol (4 atoms)<br /> - Molar mass of FePO4 = 55.85 + 30.97 + (16.00 * 4) = 150.91 g/mol<br /><br />Now, let's calculate the number of moles of each reactant:<br /><br />1. Moles of calcium carbonate (CaCO3):<br /> - Moles = Mass / Molar mass<br /> - Moles of CaCO3 = 50 g / 100.09 g/mol = 0.500 mol<br /><br />2. Moles of Iron (III) phosphate (FePO4):<br /> - Moles = Mass / Molar mass<br /> - Moles of FePO4 = 15 g / 150.91 g/mol = 0.099 mol<br /><br />Next, we need to compare the mole ratio of the reactants to the stoichiometry of the balanced chemical equation:<br /><br />3CaCO3 + 2FePO4 → Ca3(PO4)2 + Fe2(CO3)3<br /><br />From the balanced equation, we can see that the mole ratio of CaCO3 to FePO4 is 3:2.<br /><br />Now, let's compare the mole ratio of the reactants:<br /><br />- Mole ratio of CaCO3 to FePO4 = 0.500 mol / 0.099 mol = 5.05<br /><br />Since the mole ratio of CaCO3 to FePO4 is greater than the stoichiometric ratio of 3:2, this means that Iron (III) phosphate (FePO4) is the limiting reagent.<br /><br />Therefore, the limiting reagent in this reaction is Iron (III) phosphate (FePO4).