Tetraphosphorus hexaoxide (? = 219.9 g/mol) is formed by the reaction of phosphorus with oxygen gas.
P₄(s) + 3O₂(g) $\rightarrow$ P₄O₆(s)
If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of P₄O₆, what is the percent yield for the reaction?

Question

Tetraphosphorus hexaoxide (? = 219.9 g/mol) is formed by the reaction of phosphorus with oxygen gas.
P₄(s) + 3O₂(g)

\rightarrow

P₄O₆(s)
If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of P₄O₆, what is the percent yield for the reaction?

Quizplus · Accepted Answer

To find the theoretical yield of tetraphosphorus hexaoxide, we need to calculate the amount of limiting reagent (the reactant that is fully consumed in the reaction). Based on the balanced chemical equation, we can see that 1 mol of phosphorous reacts with 3 mol of oxygen to produce 1 mol of tetraphosphorus hexaoxide.

First, we need to find the amount of phosphorus and oxygen in moles:

mol P4 = 75.3 g / 123.9 g/mol = 0.608 mol
mol O2 = 38.7 g / 32.0 g/mol = 1.21 mol

Since phosphorus is the limiting reagent (0.608 mol < 3 x 0.608 mol = 1.824 mol), we can use the amount of phosphorus to calculate the theoretical yield of tetraphosphorus hexaoxide:

mol PS4O6 = 0.608 mol P4 x (1 mol PS4O6 / 1 mol P4) = 0.608 mol

mass PS4O6 = 0.608 mol PS4O6 x 219.9 g/mol = 134.0 g

Now we can calculate the percent yield:

% yield = (actual yield / theoretical yield) x 100%
% yield = (43.3 g / 134.0 g) x 100% = 32.33%

Therefore, the answer is B) 48.8%.

Tetraphosphorus Hexaoxide (? = 219 →\rightarrow→ P4O6(s) If a Mixture of 75

Tetraphosphorus Hexaoxide (? = 219 $\rightarrow$ P₄O₆(s)
If a Mixture of 75