How many Milliters of 18.0M H2SO4 are required to react with 250mL of 2.50 M Al(OH)3 if the products are aluminum sulfate and water?
Write the equation and balance it.
Calculate moles Al(OH)3. Moles = M x L.
Using the coefficients in the balanced equation, convert moles Al(OH)3 to moles H2SO4.
Then M H2SO4 = moles H2SO4/L H2SO4.
You know M and moles, calculate L H2SO4 and convert to mL.
Post your work if you get stuck.
Thanks!
How many millitres of 18.0MH2SO4 would be required to react with250Ml of 2.5Me Al(OH)3?
To calculate the volume of H2SO4 required for the reaction, we will first write the balanced chemical equation for the reaction between Al(OH)3 and H2SO4:
2 Al(OH)3 + 3 H2SO4 -> Al2(SO4)3 + 6 H2O
From the balanced equation, we can see that the stoichiometric ratio between Al(OH)3 and H2SO4 is 2:3. This means that 2 molecules of Al(OH)3 react with 3 molecules of H2SO4.
Now let's calculate the number of moles of Al(OH)3 in 250 mL of 2.50 M solution:
Molarity (M) = moles/volume (L)
2.50 M = moles/0.250 L
Rearranging the equation, we find:
moles = 2.50 M x 0.250 L = 0.625 moles
Since the stoichiometry between Al(OH)3 and H2SO4 is 2:3, we need 0.625 moles of Al(OH)3 to react with (3/2) x 0.625 = 0.938 moles of H2SO4.
Now let's find the volume of 18.0 M H2SO4 needed to obtain 0.938 moles:
Molarity (M) = moles/volume (L)
18.0 M = 0.938 moles/volume (L)
Rearranging the equation:
volume (L) = 0.938 moles / 18.0 M = 0.0521 L
Finally, convert the volume from liters to milliliters:
0.0521 L x 1000 mL/L = 52.1 mL
Therefore, 52.1 mL of 18.0 M H2SO4 are required to react with 250 mL of 2.50 M Al(OH)3.