Determine the limiting reactant for the reaction, Calculate the mass of H3PO4 produced in the reaction, Calculate the percent yield of H3PO4 if 45.7 g H3PO4 is isolated.

Question

A sample of 78.3 g78.3 g of tetraphosphorous decoxide (P4O10)(P4⁢O10) reacts with 78.1 g78.1 g of water to produce phosphoric acid (H3PO4)(H3⁢PO4) according to the following balanced equation.

P4O10+6H2O⟶4H3PO4

Answer 1

The molecular mass of P4O10 is 283.9 g/mole,  the molecular mass of H3PO4 is 97.99 g/mole, and the molecular mass of H2O is 18.02 g/mole.

The number of moles of P4O10 is 78.3 / 283.9 = 0.276 moles, and the number of moles of H2O is 78.1 / 18.02 = 4.33 moles.  The number of moles of H3PO4 produced is 45.7 / 97.99 = 0.466 moles.  

According to the balanced equation, if H2O is in excess, 0276 moles of P4O10 should produce 4  x 0.276 = 1.104 moles of H3PO4. As only 0.466 moles of H3PO4 were produced, the P4O10 is the limiting reactant.  On the other hand, if P4O10 is in excess, you would need 0.120 moles of H2O, so H2O is in excess.

Since only 45.7 g of H3PO4 were produced, but the expected amount would be 1.104 moles which is equivalent to 1.104 moles x 97.99 g/mole = 108.2g, the percent yield is 45.7g / 108.2 g x 100 = 42.2 %.