Question
How many mL of 18 M H2SO4 are required to react with 250 mL of 2.50 M Al(OH)3 if the products are aluminum sulfate and water? How do i solve this?
Answers
DrBob222
1.Write the equation and balance it.
2.Convert 250 mL of 2.50 Al(OH)3 to moles. moles = M x L.
3. Using the coefficients in the balanced equation, convert moles Al(OH)3 to moles H2SO4.
4. Convert moles H2SO4 to volume. moles = M x L. Solve for L and convert to mL.
2.Convert 250 mL of 2.50 Al(OH)3 to moles. moles = M x L.
3. Using the coefficients in the balanced equation, convert moles Al(OH)3 to moles H2SO4.
4. Convert moles H2SO4 to volume. moles = M x L. Solve for L and convert to mL.
becca
i. calculate the pH of the .833 sodium acetate solution
ii. Calculate the pH after adding .700 grams of NaOH to the 35 ml of .833 M sodium acetate solution
iii. Calculate the pH after adding 1.5 mL of 12 M HCl to the original 35 mL of .833 M sodium acetate solution
ii. Calculate the pH after adding .700 grams of NaOH to the 35 ml of .833 M sodium acetate solution
iii. Calculate the pH after adding 1.5 mL of 12 M HCl to the original 35 mL of .833 M sodium acetate solution
David
250mL x 1L/1000mL x 2.50mol Al(OH)3/1L x 3 mol H2SO4/2 mol Al(OH)3 x 1L/18.0 mol H2SO4 x 1000mL/1L = 52.1mL H2SO4