How many moles and how many grams of potassium alum, KAl(SO4)2 · 12 H2O, can be prepared from 5.00 g of aluminum?
To determine the number of moles and grams of potassium alum that can be prepared from 5.00 g of aluminum, we need to use stoichiometry.
Stoichiometry is the calculation of the quantities of reactants and products in a chemical reaction. It involves using balanced chemical equations to determine the mole ratios between the substances involved.
To begin, we need to write the balanced chemical equation for the reaction in which aluminum reacts with potassium sulfate to form potassium alum:
2 Al + 3 K2SO4 + 12 H2O -> 2 KAl(SO4)2 · 12 H2O + 3 H2
According to the balanced equation, we can see that 2 moles of aluminum react with 3 moles of potassium sulfate to produce 2 moles of potassium alum. This gives us a mole ratio of 2:3 between aluminum and potassium alum.
Now, we can calculate the number of moles of aluminum in 5.00 grams using its molar mass. The molar mass of aluminum is 26.98 g/mol.
Number of moles of aluminum = mass of aluminum / molar mass of aluminum
= 5.00 g / 26.98 g/mol
= 0.1852 mol
Since the mole ratio between aluminum and potassium alum is 2:3, we can now calculate the number of moles of potassium alum that can be prepared:
Number of moles of potassium alum = Number of moles of aluminum * (2 moles of KAl(SO4)2 · 12 H2O / 2 moles of Al)
= 0.1852 mol * (2 mol / 2 mol)
= 0.1852 mol
Therefore, 0.1852 moles of potassium alum can be prepared from 5.00 grams of aluminum.
To calculate the mass of potassium alum, we use its molar mass. The molar mass of potassium alum is 474.39 g/mol.
Mass of potassium alum = Number of moles of potassium alum * molar mass of potassium alum
= 0.1852 mol * 474.39 g/mol
= 87.89 g
Therefore, 87.89 grams of potassium alum can be prepared from 5.00 grams of aluminum.