Answer:
3 e⁻ transfer has occurred.
Explanation
This is a redox reaction.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet or duplet configuration. An octet configuration is that of outer shell configuration of noble gas.
- [Ne]= (1s²) (2s² 2p⁶)
A combination of both the reactions( Half-reactions) leads to a redox reaction.
Let us look at initial configurations of Al and Cl
[Al]= 1s² 2s² 2p⁶ 3s² 3p¹
[Cl]= 1s² 2s² 2p⁶ 3s² 3p⁵
Hence, Al can lose 3 electrons to achieve octet config.
and, Cl can gain 1e to achieve nearest noble gas config. [Ar]
This reaction can be rewritten, by clearly mentioning the oxidation states of all the entities involved.
Al⁰ + Cl⁰ → (Al⁺³)(Cl⁻)₃
Here, Aluminum is undergoing an oxidation(i.e loss of electrons) from: 0→(+3)
Chlorine undergoes a reduction half reaction (i.e gain of electrons) from: 0→(-1). There are 3 such chlorine atoms, hence 3 e⁻ transfer has occurred.
Most elements on group 18 are the Noble Gases. They already have a complete last level with 8 electrones. Actually they can form compounds but only on the lab and they will not even last half a second.
Answer:
There is a production of 11.6 moles of CO₂
Explanation:
The reaction is this:
2C₂H₆(g) + 7O₂(g) ⟶ 4CO₂(g) + 6H₂O(g)
2 moles of ethane reacts with 7 moles of oxygen, to make 4 mol of dioxide and 6 moles of water vapor.
If the oxygen is in excess, we make the calculate with the ethane (limiting reactant)
2 moles of ethane produce 4 moles of dioxide
5.8 moles of ethane produce (5.8 .4)/2 = 11.6 moles
Answer:
Unlike isopropanol, hydrogen peroxide is not a type of alcohol. You might recognize its chemical formula, H2O2, as being similar to that of water (H2O). The difference is that hydrogen peroxide has two oxygen atoms instead of one. That one extra oxygen atom makes it a strong oxidizer.
Balanced equation:
<span>2 NO + 5 H2 ------> 2 NH3 + 2 H2O
</span>
<span>2 moles NO react with 5 moles H2 to produce 2 moles NH3
</span>
<span>Molar mass of NO = 30.00 g/mol </span>
<span>86.3g NO = 86.3/30.00 = 2.877 moles of NO </span>
<span>This will require: 2.877*5 / 2 = 7.192 moles of H2 </span>
<span>Molar mass of H2 = 2 g/mol </span>
<span>25.6g H2 = 25.6/2 = 12.7 mol H2. </span>
<span>You have excess H2 means the NO is limiting </span>
<span>From the balanced equation: </span>
<span>2 moles of NO will produce 2 moles of NH3 </span>
<span>2.877 moles of NO will produce 2.877 moles of NH3 </span>
<span>Molar mass NH3 = 17g/mol </span>
<span>Mass NH3 produced = 2.877 * 17 = 48.91g
Hence the yield is = 48.91 g ~ 49 g</span>
