Answer:
Limiting reactant is NiSO₄
Explanation:
The reaction of aluminum metal with aqueous nickel(II) sulfate to produce aqueous aluminum sulfate and nickel is:
2 Al(s) + 3 NiSO₄ → Al₂(SO₄)₃ + 3 Ni
<em>That means 2 moles of Al react with 3 moles of nickel sulfate.</em>
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Moles of Al and NiSO₄ are:
Al: 108g × (1mol / 26.98g) = 4.00 moles of Al
NiSO₄: 464g × (1mol / 154.75g) = 3.00 moles of NiSO₄
For a complete reaction of aluminium there are necessary:
4.00mol Al ₓ ( 3 moles NiSO₄ / 2 moles Al) = 6 moles of NiSO₄
As you have just 3.00 moles of NiSO₄, the <em>limiting reactant is NiSO₄</em>
File:///Users/user/Downloads/Chemistry%20Exam%20Study%20Guide%20Fall%202012.pdf
Idk if this will help you, cuz it says what question are going to be in ur final
Answer:
a.
ZnS(s) + O2(g) → ZnO(s) + SO2(g)
b.
HCl (aq) + Ba(OH)2(aq) → BaCl2(aq) + H2O
Answer:
When the coefficients in a balanced chemical reaction are multiplied by two, the equilibrium constant is not affected.
Explanation:
The equilibrium constant of a reaction is known to remain steady.
Even if all the coefficients of all the species in the reaction are multiplied by two, the value of the equilibrium constant will reamin the same because the equilibrium position will not change as a result of that.
Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.