22.5 grams of lithium reacts with 33.5 grams of aqueous aluminum sulfate
aND?
To determine the products of the reaction between lithium and aqueous aluminum sulfate, we first need to write out the balanced chemical equation for the reaction.
The chemical formula for lithium is Li, and the chemical formula for aluminum sulfate is Al2(SO4)3.
The balanced chemical equation for the reaction can be written as:
2 Li + Al2(SO4)3 -> 3 Li2SO4 + 2 Al
Now that we have the balanced equation, we can calculate the amount of products that will be formed based on the given quantities of reactants.
The molar mass of lithium (Li) is approximately 6.94 g/mol, while the molar mass of aluminum sulfate (Al2(SO4)3) is approximately 342.15 g/mol.
First, let's calculate the number of moles of lithium (Li) and aqueous aluminum sulfate (Al2(SO4)3) based on their masses:
Number of moles of Li = mass of Li (in grams) / molar mass of Li
= 22.5 g / 6.94 g/mol
= 3.24 mol (approximately)
Number of moles of Al2(SO4)3 = mass of Al2(SO4)3 (in grams) / molar mass of Al2(SO4)3
= 33.5 g / 342.15 g/mol
= 0.098 mol (approximately)
According to the balanced chemical equation, the mole ratio between Li and Al2(SO4)3 is 2:1. This means that for every 2 moles of Li, we need 1 mole of Al2(SO4)3 to completely react.
Comparing the moles of Li and Al2(SO4)3, we can see that there are excess moles of Li:
Excess moles of Li = moles of Li - (2 * moles of Al2(SO4)3)
= 3.24 mol - (2 * 0.098 mol)
= 3.24 mol - 0.196 mol
= 3.04 mol
This means that 3.04 moles of Li will be consumed in the reaction, and the remaining 0.20 moles of Li will be in excess.
Now, let's calculate the moles of each product formed:
The mole ratio between Li2SO4 and Li is 3:2. This means that for every 3 moles of Li, we will get 2 moles of Li2SO4.
Moles of Li2SO4 = (3 / 2) * excess moles of Li
= (3 / 2) * 3.04 mol
= 4.56 mol
Therefore, the reaction between 22.5 grams of Li and 33.5 grams of aqueous Al2(SO4)3 will produce approximately 4.56 moles of Li2SO4.