How many liters of oxygen are required to react completely with 14.8 mol of Al?
1 answer:
Answer:
296 L
Explanation:
We will need a balanced equation with moles, so let's gather all the information in one place.
4Al + 3O₂ ⟶ 2Al₂O₃
n/mol: 17.4
1. Moles of O₂
2. Volume of O₂
You haven't given the conditions at which the volume is measured, so I assume it is at STP (0 °C and 1 bar).
At STP the molar volume of a gas is 22.71 L.
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Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂
(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.
(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed
Taking into account the definition of calorimetry and latent heat, a heat of 2159.5 J is needed to melt 35 g of iodine.
<h3>Calorimetry</h3>Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
<h3>Latent heat</h3>Latent heat is defined as the energy required by a quantity of substance to change state.
When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.
The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to
Q = m×L
where L is called the latent heat of the substance and depends on the type of phase change.
<h3>Heat needed to melt iodine</h3>In this case, you know:
- m= 35 g
- L=61.7
Replacing in the definition of latent heat:
Q= 35 g× 61.7
Solving:
<u><em>Q=2159.5 J</em></u>
Finally, a heat of 2159.5 J is needed to melt 35 g of iodine.
Learn more about calorimetry:
#SPJ1
Answer:
Sr(OH)₂ will be the limiting reagent.
Explanation:
First of all, you should know the following balanced chemical equation:
2 H₂CO₃ + 2 Sr(OH)₂ → 4 H₂O + Sr₂(CO₃)₂
The balanced equation is based on the Law of Conservation of Mass, which says that matter cannot be created or destroyed. Therefore, the number of each type of atom on each side of a chemical equation must be the same.
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). By stoichiometry the following amounts in moles react:
- strontium hydroxide: 2 moles
- carbonic acid: 2 moles
Now, you know the following masses of the elements:
- Sr: 87.62 g/mole
- O: 16 g/mole
- H: 1 g/mole
So the molar mass of strontium hydroxide is:
Sr(OH)₂= 87.62 g/mole + 2*(16 g/mole + 1 g/mole)= 121.62 g/mole
You apply the following rule of three, if 121.62 grams of hydroxide are present in 1 mole, 12.5 grams in how many moles are they?
moles of hydroxide= 0.103 moles
On the other hand, you have 150 ml of 3.5 M carbonic acid. Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you can apply the following rule of three: if in 1 L there are 3.5 moles of carbonic acid, in 0.150 L (being 1 L = 1000 mL, 0.150 L = 150 mL) how many moles of acid are there?
moles of carbonic acid= 0.525 moles
Finally, to calculate the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 2 mole of strontium hydroxide reacts with , how much moles of carbonic acid will be needed if 0.103 moles of strontium hydroxide react?
moles of carbonic acid= 0.103 moles
But 0.525 moles are available. Since more moles are available than you need to react with 0.103 moles of strontium hydroxide, <u><em>Sr(OH)₂ will be the limiting reagent.</em></u>