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%
The heat of solution is -51.8 kJ/mol
<h3>What is the heat of solution?</h3>
We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.
Number of moles of KOH = 11.9-g/56 g/mol = 0.21 moles
Temperature rise = 26.0 ∘c
Mass of the water = 100.0 grams
Heat capacity = 4.184 j/g⋅°c
Then;
ΔH = mcθ
ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ
Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol
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Answer: 16.32 g of 
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of 
Thus 16.32 g of as excess reagent are left.
