What volume of H2SO4 0.04 M is required to completely neutralize 200 mL of Al(OH)3 0.02M?
To determine the volume of H2SO4 0.04 M required to neutralize 200 mL of Al(OH)3 0.02M, you will need to use the balanced chemical equation and perform a stoichiometric calculation.
The balanced chemical equation for the reaction between H2SO4 and Al(OH)3 is:
2 Al(OH)3 + 3 H2SO4 --> Al2(SO4)3 + 6 H2O
From the balanced equation, we can see that the mole ratio between Al(OH)3 and H2SO4 is 2:3. This means that for every 2 moles of Al(OH)3, we need 3 moles of H2SO4.
First, let's calculate the moles of Al(OH)3 in 200 mL of a 0.02 M solution:
Moles of Al(OH)3 = concentration (M) x volume (L)
= 0.02 mol/L x 0.2 L
= 0.004 mol
Since the mole ratio between Al(OH)3 and H2SO4 is 2:3, we can calculate the moles of H2SO4 required:
Moles of H2SO4 = (moles of Al(OH)3) x (3 moles of H2SO4 / 2 moles of Al(OH)3)
= 0.004 mol x (3/2)
= 0.006 mol
Now, we can calculate the volume of H2SO4:
Volume of H2SO4 = (moles of H2SO4) / (concentration of H2SO4)
= 0.006 mol / 0.04 mol/L
= 0.15 L or 150 mL
Therefore, 150 mL of H2SO4 0.04 M is required to completely neutralize 200 mL of Al(OH)3 0.02 M.
To find the volume of H2SO4 required to neutralize Al(OH)3, we can use the concept of stoichiometry.
First, let's write the balanced chemical equation for the neutralization reaction between H2SO4 and Al(OH)3:
2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O
From the balanced equation, we can see that 2 moles of Al(OH)3 react with 3 moles of H2SO4.
Given the concentration of Al(OH)3 is 0.02 M and the volume is 200 mL, we need to convert the volume into liters:
200 mL = 200 / 1000 = 0.2 L
To find the number of moles of Al(OH)3, we multiply the concentration by the volume:
Moles of Al(OH)3 = concentration × volume
= 0.02 M × 0.2 L
= 0.004 moles
According to the balanced equation, it takes 3 moles of H2SO4 to react with 2 moles of Al(OH)3.
Therefore, the moles of H2SO4 required would be:
Moles of H2SO4 = (moles of Al(OH)3 × 3) / 2
= (0.004 moles × 3) / 2
= 0.006 moles
Since the concentration of H2SO4 is 0.04 M, we can now find the volume of H2SO4 required:
Volume of H2SO4 = moles of H2SO4 / concentration
= 0.006 moles / 0.04 M
= 0.15 L or 150 mL
Therefore, 150 mL of H2SO4 0.04 M is required to completely neutralize 200 mL of Al(OH)3 0.02 M.
first construct the balance equation for sulphuric acid and aluminium hydroxide reaction.
take note of the mole ratio between the two compound from your balanced equation.
determine the actual mole of Al(OH)3 using c=n/v where c is the concentration in mol/L, n is the unknown mole, and v is volume in mL
use the mole ratio to determine the mole of sulphuric acid
calculate the volume of sulphuric acid using the same formula c=n/v
hope that helps