Phosphoric acid, H3PO4 is one of the main ingredients of soft drinks, detergents, and fertilizers. It can be prepared with a series of reactions: P4 + 5O2 → P4O10 P4O10+6H2O → 4H3PO4Let us say we allow 1000 kg of phosphorus to react with oxygen in the tank to yield 90% of tetra phosphorus decaoxide (P4O10). In the second step of the reaction, we react it with water to yield 97% of H3PO4. How much, in kilogram, of H3PO4 was produced after a series of reactions?

Explanation

To calculate the amount of H3PO4 produced, we need to first find the amount of tetraphosphorus decaoxide (P4O10) produced and then the amount of H3PO4 produced from it. Molecular weight of P4: 4(31) = 124 kg/mol Molecular weight of H3PO4: 3(1) + 1(31) + 4(16) = 98 kg/mol From the given reaction equation, we know that: 1 kmol of P4 produces 1 kmol of P4O10 (124 kg) So, 1000 kg of P4 will produce: 1000 kg P4 / 124 kg P4/kmol = 8.064516129 kmol P4 Since the actual yield of P4O10 is 90%, the actual amount of P4O10 produced is: 0.9 x 8.064516129 kmol = 7.258064516 kmol From the second reaction equation, we know that: 1 kmol of P4O10 produces 4 kmol of H3PO4 (98 kg) So, 7.258064516 kmol of P4010 will produce: 7.258064516 kmol x 4 kmol H3PO4/kmol P4O10 x 98 kg H3PO4/kmol H3PO4 = 2,859.67741935 kg H3PO4 Since the actual yield of H3PO4 is 97%, the actual amount of H3PO4 produced is: 0.97 x 2,859.67741935 kg = 2,759.81 kg Therefore, the answer is C) 2759.81 kg of H3PO4 was produced.