Question

A 80 mL sample of sulfuric acid solution is neutralize by 254.4 mL of a 1.3 M aluminum hydroxide solution. What is the molarity of the acid?

Answer 1

Is it 4.14?

Answer 2

No. See the problem below it that I worked for you.
3H2SO4 + 2Al(OH)3 ==> Al2(SO4)3 + 6H2O
Post your work if you get stuck and explain what is giving you trouble.

Answer 3

H2SO4 + 2NaOH = 2H2O + Na2SO4
.2545 L NaOH x 1.3M NaOH = .33085 mol NaOH
.33085 mol NaOH x (1 mol H2SO4/2 mol NaOH) / .080L H2SO4=4.14

Answer 4

Ugh. I used the wrong stuff and did something with sodium

Answer 5

3H2SO4 + 2Al(OH)3 ==> Al2(SO4)3 + 6H2O
.2545 L x 1.3M 2Al(OH)3 = .33085 mol
.33085 mol x (1 mol H2SO4/2 mol ) / .080L H2SO4=4.14

Answer 6

Step 1. Write and balance the equation. That's done.
Step 2. Calculate moles of what you have. That's moles Al(OH)3 = M x L = 0.2545 x 1.3 = 0.3308
Step 3. Convert mols Al(OH)3 to moles H2SO4 using the coefficients in the balanced equation.
0.3308 mols Al(OH)3 x [3 moles H2SO4/2 mols Al(OH)3] =
0.3308 x 3/2 = 0.4962
Step 4. Then M H2SO4 = moles H2SO4/L H2SO4 = 0.4962/0.080 = 6.20 M
Look at this step by step procedure. It will work titration problems but also most stoichiometry problems. Follow up with questions here please.