Find the volume of 0.110 M hydrochloric acid necessary to react completely with 1.54 Al(OH)3.
Please explain how to do this.
1.54 what? tons, buckets, grams?
To find the volume of hydrochloric acid necessary to react completely with 1.54 Al(OH)3, you need to use stoichiometry and the balanced chemical equation of the reaction.
First, let's write the balanced chemical equation for the reaction between hydrochloric acid (HCl) and aluminum hydroxide (Al(OH)3):
3HCl + Al(OH)3 -> AlCl3 + 3H2O
From the balanced chemical equation, we can see that 3 moles of HCl react with 1 mole of Al(OH)3.
Next, we need to convert the given concentration of hydrochloric acid to moles per liter (Molarity) using the formula: concentration (M) = moles/volume (L).
Given that the concentration of HCl is 0.110 M, we can assume that 0.110 moles of HCl are present in 1 L (1000 mL) of the solution.
Using the mole-to-mole ratio from the balanced equation, we can determine the number of moles of Al(OH)3 present.
1.54 moles of Al(OH)3 x (3 moles of HCl / 1 mole of Al(OH)3) = 4.62 moles of HCl
Since the mole ratio is 3:1 (HCl:Al(OH)3), we know that 4.62 moles of HCl are required to react completely with 1.54 moles of Al(OH)3.
Finally, to find the volume of 0.110 M HCl required, we use the formula:
moles = concentration x volume
Rearranging the formula, we can solve for volume:
volume = moles / concentration
volume = 4.62 moles / 0.110 M
Calculating this, we find:
volume = 42 mL
Therefore, 42 mL of 0.110 M hydrochloric acid is required to react completely with 1.54 moles of aluminum hydroxide.
To find the volume of hydrochloric acid (HCl) necessary to react completely with aluminum hydroxide (Al(OH)3), you need to use the balanced chemical equation for the reaction.
First, write the balanced chemical equation for the reaction between hydrochloric acid and aluminum hydroxide:
2Al(OH)3 + 6HCl -> 2AlCl3 + 6H2O
According to the balanced equation, two moles of aluminum hydroxide react with six moles of hydrochloric acid to produce two moles of aluminum chloride and six moles of water.
Next, convert the given concentration of the hydrochloric acid into moles per liter (Molarity or M):
0.110 M hydrochloric acid means there are 0.110 moles of HCl per liter of solution.
Now, determine the number of moles of hydrochloric acid needed to react completely with the given amount of aluminum hydroxide:
From the balanced equation, we see that 2 moles of aluminum hydroxide react with 6 moles of hydrochloric acid. Therefore, the molar ratio is 2:6, which simplifies to 1:3.
So, for every 1.54 moles of aluminum hydroxide, you would need 3 times that amount of hydrochloric acid (since the ratio is 1:3).
1.54 moles Al(OH)3 × 3 moles HCl / 2 moles Al(OH)3 = 2.31 moles HCl
Now, convert the moles of hydrochloric acid into volume using its given concentration:
Volume (L) = Moles / Molarity
Volume (L) = 2.31 moles / 0.110 M = 21.0 L
Therefore, the volume of 0.110 M hydrochloric acid necessary to react completely with 1.54 moles of aluminum hydroxide is 21.0 liters.