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?
6 answers
Is it 4.14?
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.
3H2SO4 + 2Al(OH)3 ==> Al2(SO4)3 + 6H2O
Post your work if you get stuck and explain what is giving you trouble.
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
.2545 L NaOH x 1.3M NaOH = .33085 mol NaOH
.33085 mol NaOH x (1 mol H2SO4/2 mol NaOH) / .080L H2SO4=4.14
Ugh. I used the wrong stuff and did something with sodium
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
.2545 L x 1.3M 2Al(OH)3 = .33085 mol
.33085 mol x (1 mol H2SO4/2 mol ) / .080L H2SO4=4.14
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.
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.