Explanation:
Fe. O
72.40/ 56 27.60/16
____________________
1.29/1.29 1.725/1.29
_____________________
1. :. 1
<h3>Emperical formula = FeO</h3>
Answer:
Ammonia is limiting reactant
Amount of oxygen left = 0.035 mol
Explanation:
Masa of ammonia = 2.00 g
Mass of oxygen = 4.00 g
Which is limiting reactant = ?
Balance chemical equation:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 2.00 g/ 17 g/mol
Number of moles = 0.12 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 4.00 g/ 32 g/mol
Number of moles = 0.125 mol
Now we will compare the moles of ammonia and oxygen with water and nitrogen.
NH₃ : N₂
4 : 2
0.12 : 2/4×0.12 = 0.06
NH₃ : H₂O
4 : 6
0.12 : 6/4×0.12 = 0.18
O₂ : N₂
3 : 2
0.125 : 2/3×0.125 = 0.08
O₂ : H₂O
3 : 6
0.125 : 6/3×0.125 = 0.25
The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.
Amount of oxygen left:
NH₃ : O₂
4 : 3
0.12 : 3/4×0.12= 0.09
Amount of oxygen react = 0.09 mol
Amount of oxygen left = 0.125 - 0.09 = 0.035 mol
Al
Explanation:
The limiting reactant will be Al:
4Al + 3O₂ → 2Al₂O₃
The limiting reactant is the reactant in short supply in a chemical reaction.
Given parameters:
Mass of Al = 30g Molar mass = 27g/mol
Number of moles = 
= 
Number of moles of Al = 1.111 mole
Mass of O₂ = 30g, molar mass = 32g/mol
Number of moles = 
= 0.94mol
In the reaction:
4 moles of Al reacted with 3 moles of O₂
1.11moles of Al will require 
= 0.83mole to react
But we have been given 0.94mole of O₂. This is more than required.
Therefore O₂ is in excess and Al is the limiting reactant.
Learn more:
Limiting reagents brainly.com/question/6078553
#learnwithBrainly
Answer:
a. 0.393M CH₃COOH.
b. 2.360% of acetic acid in the solution
Explanation:
The reaction of acetic acid (CH₃COOH) with NaOH is:
CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺
<em>That means 1 mole of acid reacts per mole of NaOH.</em>
Moles of NaOH to reach the equivalence point are:
35.75mL = 0.03575L × (0.2750mol / L) = <em>9.831x10⁻³ moles of NaOH</em>
As 1 mole of acid reacts per mole of NaOH, moles of CH₃COOH in the acid solution are 9.831x10⁻³ moles.
a. As the volume of the acetic acid solution is 25.00mL = 0.02500L, the molarity of the solution is:
9.831x10⁻³ moles / 0.02500L =
<h3>0.393M CH₃COOH</h3>
b. Molar mass of acetic acid is 60g/mol. The mass of 9.831x10⁻³ moles is:
9.831x10⁻³ moles ₓ (60g / mol) = <em>0.590g of CH₃COOH</em>.
As volume of the solution is 25.00mL, the percentage of acetic acid is:
(0.590g CH₃COOH / 25.00mL) ₓ 100 =
<h3>2.360% of acetic acid in the solution</h3>
