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.First, let's calculate the molar masses of the reactants and products:1. **Calcium Carbonate ( )**: - - - - Total: 2. **Iron(III) Phosphate ( )**: - - - - Total: 3. **Iron(III) Carbonate (\(Fe_2(CO_3)_3\))**: - - - - Total: Next, we convert the masses of the reactants to moles:1. **Calcium Carbonate**: 2. **Iron(III) Phosphate**: Now, we use the stoichiometry of the balanced equation to determine the limiting reactant: From the equation, 3 moles of react with 2 moles of . Therefore, the ratio of to should be 3:2.We need to check which reactant is the limiting reactant by comparing the actual mole ratio to the stoichiometric ratio: Since the actual ratio (5.05) is greater than the stoichiometric ratio (1.5), is the limiting reactant.Now, we calculate the maximum amount of \(Fe_2(CO_3)_3\) that can be produced from the limiting reactant: Finally, we convert the moles of \(Fe_2(CO_3)_3\) to grams:\[\text{Mass of } Fe_2(CO_3)_3 = 0.0495 \, \text{mol} \times