Answer to Question #313607 in General Chemistry for Ivan Ackerman

Question

A sample of 94.9 g

94.9 g of tetraphosphorous decoxide (P4⁢O10) reacts with 68.0 g

68.0 g of water to produce phosphoric acid (H3⁢PO4) according to the following balanced equation.

P4O10+6H2O⟶4H3PO4

Determine the limiting reactant for the reaction.

Calculate the mass of H3⁢PO4 produced in the reaction.

Calculate the percent yield of H3⁢PO4 is isolated after carrying out the reaction.

Answer 1

Balanced equation is given as follows

$P_4O_{10} +6H_2O \longrightarrow 4H_3PO_4$

Mass of $P_4O_{10}$ = 94.9 g

Mass of H₂O = 68 g

Mole of P₄O₁₀ = $\frac{94.9}{283.88}= 0.334$

Mole of $H_2O= \frac{68}{18}=3.78$

1 mole of P₄O₁₀ reacts with 6 mole of H₂O to give 4 mole of H₃PO₄

so, 0.334 mole of P₄O₁₀ reacts with 2.004 mole of H₂O

so here limiting reagent is P₄O₁₀

mole of H₃PO₄ produces = 1.336 mole

mass of H₃PO₄ produces = 130.9 g

% yield of the reaction is $\frac{1.336 }{4}\times 100=33.4$ %