Aluminium react with oxygen to form aluminium oxide.how many grams of potassium chlorate V would be heated to produce enough oxygen to form 5.1grams of aluminium oxide
25
To determine how many grams of potassium chlorate (KClO3) are required to produce enough oxygen to form 5.1 grams of aluminum oxide (Al2O3), we need to consider the balanced chemical equation for the reaction:
4Al + 3O2 → 2Al2O3
From the equation, we can see that 3 moles of oxygen (O2) are required to produce 2 moles of aluminum oxide (Al2O3). Therefore, we need to find the molar mass of Al2O3 and convert it to moles.
1 mole of Al2O3 = molar mass of Al2O3 (g)
The molar mass of Al2O3 is calculated as follows:
2(Al) + 3(O) = 2(26.98 g/mol) + 3(16.00 g/mol) = 101.96 g/mol
Now, we can calculate the moles of Al2O3 required:
moles of Al2O3 = mass of Al2O3 (g) / molar mass of Al2O3 (g/mol)
moles of Al2O3 = 5.1 g / 101.96 g/mol ≈ 0.05 mol
Since 2 moles of Al2O3 require 3 moles of O2, we can calculate the moles of O2 needed:
moles of O2 = 0.05 mol Al2O3 × (3 mol O2 / 2 mol Al2O3) ≈ 0.075 mol
Finally, we can determine the mass of potassium chlorate required to produce this amount of oxygen. The molar mass of KClO3 is:
K = 39.10 g/mol
Cl = 35.45 g/mol
O = 16.00 g/mol
3(O) = 3(16.00 g/mol) = 48.00 g/mol
molar mass of KClO3 = K + Cl + 3(O) = 39.10 g/mol + 35.45 g/mol + 48.00 g/mol = 122.55 g/mol
mass of KClO3 required = moles of O2 × molar mass of KClO3
mass of KClO3 required = 0.075 mol × 122.55 g/mol ≈ 9.19 g
Therefore, approximately 9.19 grams of potassium chlorate (KClO3) would be needed to produce enough oxygen to form 5.1 grams of aluminum oxide (Al2O3).