aluminum hydroxide is decomposed to form aluminum oxide and water. if 2.23 moles of water are formed then how many of aluminum oxide are formed. how many grams of aluminum hydroxide must have been.decomposed?
4Al(OH)3 ==> 2Al2O3 + 6H2O
2.23 mols H2O x (2 mols Al2O3/6 mols H2O) = 2.23 x 2/6 = ? Look at the coefficients. They tell you the answer.
For part b, how many mols Al(OH)3 must you have started with. That's
2.23 mols H2O x (4 mols Al(OH)3/6 mols H2O). Then convert to grams.
g Al(OH)3 = mols x molar mass = ?
To determine the number of moles of aluminum oxide formed, we need to use the balanced chemical equation for the decomposition of aluminum hydroxide. The equation is:
2 Al(OH)3 → Al2O3 + 3 H2O
From the balanced equation, we know that for every 2 moles of aluminum hydroxide (Al(OH)3) decomposed, 1 mole of aluminum oxide (Al2O3) is formed. We are given that 2.23 moles of water (H2O) are formed.
Since the molar ratio between aluminum oxide and water is 1:3, we can determine the number of moles of aluminum oxide formed by dividing the moles of water by the molar ratio:
2.23 moles H2O ÷ 3 = 0.743 moles Al2O3
Therefore, 0.743 moles of aluminum oxide are formed.
To calculate the number of grams of aluminum hydroxide decomposed, we need to convert the moles of aluminum hydroxide to grams. To do this, we need the molar mass of aluminum hydroxide (Al(OH)3), which is:
Aluminum (Al): 1 atom x 26.98 g/mol = 26.98 g/mol
Oxygen (O): 3 atoms x 16.00 g/mol = 48.00 g/mol
Hydrogen (H): 3 atoms x 1.01 g/mol = 3.03 g/mol
Total molar mass of Al(OH)3: 26.98 g/mol + 48.00 g/mol + 3.03 g/mol = 77.01 g/mol
Now we can calculate the grams of aluminum hydroxide decomposed:
Grams of Al(OH)3 = moles of Al(OH)3 x molar mass of Al(OH)3
= 2.23 moles x 77.01 g/mol
= 171.90 grams
Therefore, 171.90 grams of aluminum hydroxide must have been decomposed.