First, let us write the reaction. We know that the reactants are Aluminum and Sulfuric acid, and one of the products is hydrogen. That means the reaction is a single replacement reaction as shown below:
2 Al + 3 H₂SO₄ = 3 H₂ + Al₂(SO₄)₃
Next, let us determine which of the reactants is limiting:
1.80 g Al (1 mol/26.98 g)(3 mol H₂SO₄/2 mol Al)(98 g H₂SO₄/mol) = 9.807 g
It would need 9.807 g. But the only available amount is 6 g. That means that H₂SO₄ is the limiting reactant. We use 6 g as the basis to know the theoretical yield:
6 g*(1 mol H₂SO₄/98 g)*(3 mol H₂/ 3 mol H₂SO₄)*(2 g /mol H₂) = 0.122 g
Therefore, the percent yield is equal to
Percent yield = (0.112 g/0.122 g)*100
Percent yield = 91.8%
I would imagine that n-butanol would have the higher boiling point because it has less branching and therefore stronger intermolecular forces between molecules.
Answer : The molarity after a reaction time of 5.00 days is, 0.109 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = 5.00 days
[A] = concentration of substance after time 't' = ?
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.110 M
Now put all the given values in above equation, we get:
![9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)](https://tex.z-dn.net/?f=9.7%5Ctimes%2010%5E%7B-6%7D%3D%5Cfrac%7B1%7D%7B5.00%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.110%29%7D%5Cright%29)
![[A]=0.109M](https://tex.z-dn.net/?f=%5BA%5D%3D0.109M)
Hence, the molarity after a reaction time of 5.00 days is, 0.109 M
Products. I’m not 100% sure because I don’t fully understand what your asking but if you are talking as in chemical reactions, the answer is a product.
