Problem 3: 3CaCO_(3)+2FePO_(4)arrow Ca_(3)(PO_(4))_(2)+Fe_(2)(CO_(3))_(3) Given: 50 grams of 1 poin 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 compare them to the stoichiometry of the balanced chemical equation.<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 masses of the reactants:<br />- Molar mass of CaCO3 = 40.08 (Ca) + 12.01 (C) + 3 × 16.00 (O) = 100.09 g/mol<br />- Molar mass of FePO4 = 55.85 (Fe) + 4 × 15.999 (O) + 30.97 (P) = 150.82 g/mol<br /><br />Now, let's calculate the number of moles of each reactant:<br />- Moles of CaCO3 = 50 g / 100.09 g/mol = 0.5 mol<br />- Moles of FePO4 = 15 g / 150.82 g/mol = 0.1 mol<br /><br />Next, we need to compare the mole ratio of the reactants to the stoichiometry of the balanced chemical equation:<br />- The balanced chemical equation is: 3CaCO3 + 2FePO4 → Ca3(PO4)2 + Fe2(CO3)3<br />- According to the equation, 3 moles of CaCO3 react with 2 moles of FePO4.<br /><br />Now, let's calculate the ratio of the moles of each reactant to the stoichiometry of the balanced chemical equation:<br />- Ratio of CaCO3 = 0.5 mol / 3 = 0.167<br />- Ratio of FePO4 = 0.1 mol / 2 = 0.05<br /><br />Since the ratio of FePO4 is smaller than the ratio of CaCO3, iron(III) phosphate (FePO4) is the limiting reagent.<br /><br />Therefore, the correct answer is: iron(III) phosphate.