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PLEASE HELP!!!
A 84.1 g sample of phosphorus reacts with 85.0 g of oxygen gas according to the following chemical equation.

4 P(s) + 5 O2(g) → 2 P₂O5

a. Find the limiting reactant.

b. How many grams of P₂O5 are produced in theory?

c. If only 123 g of P2O5 are produced, what is the percentage yield?

Sagot :

josephine169

Answer:

(a) oxygen

(b) 154g (to 3sf)

(c) 79.9% (to 3sf)

Explanation:

mass (g) = moles × Mr/Ar

note: eqn means chemical equation

(a)

moles of P = 84.1 ÷ 30.973 = 2.7152 moles

moles of O2 = 85÷2(16) = 2.65625 moles

Assuming all the moles of P is used up,

moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)

moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)

therefore there is insufficient moles of O2 and the limiting reactant is oxygen.

(b)

moles of P2O5 produced

= 2/5 (according to eqn) × 2.7152

= 1.08608moles

mass of P2O5 produced

= 1.08608 × [ 2(30.973) + 5(16) ]

= 154.164g

= approx. 154g to 3 sig. fig.

(c)

% yield = actual/theoretical yield × 100%

= 123/154 × 100%

= 79.870%

= approx. 79.9% (to 3sf)

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