Answer:
The answer to your question is 8,32 L
Explanation:
Data
Pressure 1 = 104.5 kPa
Volume 1 = 3.4 L
Pressure 2 = 42.7 kPa
Volume 2 = ?
Formula
- To solve this problem use the Boyle's law.
P1V1 = P2V2
- Solve for V2
V2 = P1V1 / P2
- Substitution
V2 = (104.5)(3.4) / 42.7
-Simplification
V2 = 355.3 / 42.7
- Result
V2 = 8.32 L
Aluminum oxide produced : = 79.152 g
<h3>Further explanation</h3>
Given
46.5g of Al
165.37g of MnO
Required
Aluminum oxide produced
Solution
Reaction
2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)
mol = mass : Ar
mol = 46.5 : 27
mol = 1.722
mol = 165.37 : 71
mol = 2.329
mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776
MnO as a limiting reactant(smaller ratio)
So mol Al₂O₃ based on MnO as a limiting reactant
From equation , mol Al₂O₃ :
= 1/3 x mol MnO
= 1/3 x 2.329
= 0.776
Mass Al₂O₃ (MW=102 g/mol) :
= 0.776 x 102
= 79.152 g
I think the answer is 98 grams
Answer:
A. N₂(g) + 3H₂(g) -----> 2NH₃ exothermic
B. S(g) + O₂(g) --------> SO₂(g) exothermic
C. 2H₂O(g) --------> 2H₂(g) + O₂(g) endothermic
D. 2F(g) ---------> F₂(g) exothermic
Explanation:
The question says predict not calculate. So you have to use your chemistry knowledge, experience and intuition.
A. N₂(g) + 3H₂(g) -----> 2NH₃ is exothermic because the Haber process gives out energy
B. S(g) + O₂(g) --------> SO₂(g) is exothermic because it is a combustion. The majority, if not all, combustion give out energy.
C. 2H₂O(g) --------> 2H₂(g) + O₂(g) is endothermic because it is the reverse reaction of the combustion of hydrogen. If the reverse reaction is exothermic then the forward reaction is endothermic
D. 2F(g) ---------> F₂(g) is exothermic because the backward reaction is endothermic. Atomisation is always an endothermic reaction so the forward reaction is exothermic
Answer:
The balanced equation is:
2 HNO3 + Mg ---> Mg(NO3)2 + H2
From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg
Moles of Mg:
Molar mass of Mg = 24 g/mol
Moles = Given mass / Molar Mass
Moles of Mg = 4.47 / 24 = 0.18 moles (approx)
Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed
Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3
Unreacted moles of HNO3 = Total Moles - Moles consumed
Unreacted moles of HNO3 = 0.64 moles (approx)
Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate
From the given options, we can see that 0.632 moles is the closest value to our answer
Therefore, 0.632 moles will remain after the reaction
