Consider the following reaction
Al(OH)3 + 3HCL ____>ALCL3
How many milliltres of 0.8 M HCL will react with 2g of AL(OH)3 ??
Here is a link if you get lost.
http://www.jiskha.com/science/chemistry/stoichiometry.html
1. moles Al(OH)3 = grams/molar mass.
2. Using the coefficients in the balanced equation, convert moles Al(OH)3 to moles HCl.
3. M HCl = moles HCl/L HCl. Solve for L HCl and convert to mL.
He analizado antes de poder ver la solución para ver los pasos
Gracias
roro
To determine how many milliliters of 0.8 M HCl will react with 2g of Al(OH)3, we need to use stoichiometry and the concept of molarity.
First, let's balance the chemical equation:
Al(OH)3 + 3HCl → AlCl3 + 3H2O
From the balanced equation, we can see that 1 mole of Al(OH)3 reacts with 3 moles of HCl.
Next, we need to convert the mass of Al(OH)3 to moles:
molar mass of Al(OH)3 = atomic mass of Al + 3 * (atomic mass of O + atomic mass of H)
= 26.98 + 3 * (16.00 + 1.01)
= 78.00 g/mol
moles of Al(OH)3 = mass of Al(OH)3 / molar mass of Al(OH)3
= 2 g / 78.00 g/mol
≈ 0.0256 mol
Since the stoichiometric ratio is 1:3 (Al(OH)3:HCl), we multiply the moles of Al(OH)3 by 3 to get the moles of HCl:
moles of HCl = moles of Al(OH)3 * 3
≈ 0.0256 mol * 3
≈ 0.0768 mol
Now, we can calculate the volume of 0.8 M HCl needed:
molarity (M) = moles / liters
0.8 M = 0.0768 mol / V (volume in liters)
V = 0.0768 mol / 0.8 M
V ≈ 0.096 L or 96 mL
Therefore, approximately 96 milliliters of 0.8 M HCl will react with 2 grams of Al(OH)3.