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)

Question

Answer 1

To solve this problem, we need to determine the limiting reactant and then calculate the maximum amount of

that can be produced.Given information:- 50.0 grams of calcium carbonate (

)- 15.0 grams of Iron (III) phosphate (

)Step 1: Calculate the molar mass of each reactant.- Molar mass of

= 100.09 g/mol- Molar mass of

= 150.82 g/molStep 2: Calculate the number of moles of each reactant.- Moles of

= 50.0 g / 100.09 g/mol = 0.500 mol- Moles of

= 15.0 g / 150.82 g/mol = 0.099 molStep 3: Determine the limiting reactant.- The balanced chemical equation is:

- The stoichiometric ratio of

to

is 3:2.- The actual ratio of

to

is 0.500 mol / 0.099 mol = 5.05.- Since the actual ratio is greater than the stoichiometric ratio,

is the limiting reactant.Step 4: Calculate the maximum amount of

that can be produced.- The balanced chemical equation shows that 2 moles of

produce 1 mole of

.- Therefore, the maximum amount of

that can be produced is 0.099 mol / 2 = 0.0495 mol.Therefore, the maximum amount of

that can be made is 0.0495 mol.

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