A solid that form from liquid in a chemical reaction is called
1 answer:
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Answer:
Explanation:
According to this. Let's analize the possible products a, b and c.
First, the problem states that we have 2 eq. of base, For this case, let's assume it's KOH. Now, As we are doing a reaction with base, means that this reaction can only take places under conditions of SN2 and E2, a fast reaction that is taking place in only 1 step.
With this in mind, let's analyze product a. This states that it has two sp hybridized carbon, in other words, a triple bond between two carbons. So the product is with no doubt, an alkyne.
Product b has only one sp hybridized carbon, which means that this carbon should cannot be an alkyne because we need two carbon atoms. The only way to have one atom of C sp hybridized, is with two double bonds, so product b would have to a alkene with two double bonds.
Product c do not have sp hybridized carbon, therefore, it only has two double bonds in two different Carbon atoms, so it's another alkene with two double bonds, but in two different atoms of carbon.
Picture attached show the product a, b and c. Hope this can help
Answer:
0.00 g
Explanation:
We have the masses of two reactants, so this is a limiting reactant problem.
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ 30.07 32.00
2CH₃CH₃ + 7O₂ ⟶ 4CO₂ + 6H₂O
Mass/g: 1.50 11.
2. Calculate the moles of each reactant
3. Calculate the moles of CO₂ we can obtain from each reactant
From ethane:
The molar ratio is 4 mol CO₂:2 mol C₂H₆
From oxygen:
The molar ratio is 4 mol CO₂:7 mol O₂
4. Identify the limiting and excess reactants
The limiting reactant is ethane, because it gives the smaller amount of CO₂.
The excess reactant is oxygen.
5. Mass of ethane left over.
Ethane is the limiting reactant. It will be completely used up.
The mass of ethane left over will be 0.00 g.