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)
Answer:
Nitrogen, Hydrogen, Oxygen, Chlorine, and Fluorine are all gases at room temperature.
Explanation:
Answer:
oxygen is limiting reactant
Explanation:
Given data:
Mass of phosphorus = 25.0 g
Mass of oxygen = 50.0 g
What is limiting reactant ?
Solution:
Chemical equation:
P₄ + 5O₂ → P₄O₁₀
Number of moles of P₄:
Number of moles = mass/molar mass
Number of moles = 25.0 g/ 123.89 g/mol
Number of moles = 0.20 mol
Number of moles of O₂:
Number of moles = mass/molar mass
Number of moles = 50.0 g/ 32 g/mol
Number of moles = 1.56 mol
now we will compare the moles of reactants with product:
P₄ : P₄O₁₀
1 : 1
0.20 : 0.20
O₂ : P₄O₁₀
5 : 1
1.56 : 1/5×1.56 = 0.312 mol
Less number of moles of product are formed by the oxygen thus it will act as limiting reactant.
Answer:
Mass of oxygen required = 19.5 moles × 16 g/mol = 312 g.
Explanation:
