Given that 12.25 g of aluminum sulfide is reacted with excess water. How many grams of aluminum hydroxide can be obtained? The molar mass of aluminum sulfide is 150.0 g and the molar mass of aluminum hydroxide is 78.0 g.
Al2S3 + 6H2O ==> 2Al(OH)3 + 3H2S
mols Al2S3 = grams/molar mass = ?
Use the coefficients in the balanced equation to convert mols Al2S3 to mols Al(OH)3.
now convert mols Al(OH)3 to grams Al(OH)3. That's g = mols x molar mass = ?
How many grams of HF will be made if you start with 29.6 grams of ammonia and fluorine is in excess?
To find the number of grams of aluminum hydroxide that can be obtained, we need to start by calculating the moles of aluminum sulfide using its molar mass.
First, we convert the mass of aluminum sulfide (12.25 g) to moles by dividing it by its molar mass (150.0 g/mol):
12.25 g / 150.0 g/mol = 0.0817 mol
Next, we use the balanced chemical equation to determine the stoichiometry of the reaction between aluminum sulfide and water. The balanced equation is:
Al2S3 + 6H2O → 2Al(OH)3 + 3H2S
From the equation, we can see that 1 mole of aluminum sulfide reacts to form 2 moles of aluminum hydroxide. Therefore, since we have 0.0817 moles of aluminum sulfide, we can calculate the moles of aluminum hydroxide formed as follows:
0.0817 mol Al2S3 × (2 mol Al(OH)3 / 1 mol Al2S3) = 0.1634 mol Al(OH)3
Finally, we convert the moles of aluminum hydroxide to grams using its molar mass:
0.1634 mol Al(OH)3 × 78.0 g/mol = 12.748 g
Therefore, approximately 12.748 grams of aluminum hydroxide can be obtained.