The answer is true.the molecule becomes reduced.
Answer:
Yes A redox reaction can be a combination reaction.
Two elements are mixed in a combination reaction to produce a single product.
Explanation:
Example: water formula
2H2 + O2 → 2H2O
Oxygen is reduced in this reaction when electrons are transferred from hydrogen to oxygen and hydrogen is also oxidized since oxygen from hydrogen accepts electrons. Oxygen is the oxidizing agent and hydrogen is the reducing agent.
2H2 + O2 → 2H2O
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:
285g of fluorine
Explanation:
To solve this problem we need to find the mass of Freon in grams. Then, with its molar mass we can find moles of freon and, as 1 mole of Freon, CCl₂F₂, contains 2 moles of fluorine, we can find moles of fluorine and its mass:
<em>Mass Freon:</em>
<em>2.00lbs * (454g / 1lb) = </em>908g of Freon
<em>Moles freon -Molar mass: 120.91g/mol- and moles of fluorine:</em>
908g of Freon * (1mol / 120.91g) =
7.5 moles of freon * (2moles Fluorine / mole Freon): 15 moles of fluorine
<em>Mass fluorine -Atomic mass: 19g/mol-:</em>
15 moles F * (19g / mol) =
<h3>285g of fluorine</h3>
