in the lab there is a spill of 47.5mL of .500 mol/L Al(OH)3. How many grams of H2SO4 would be needed to neutralize the spill.

Answer 1

Write the equation.
2Al(OH)3 + 3H2SO4 ==>Al2(SO4)3 + 6H2O

moles Al(OH)3 = M x L = 0.500 x 0.0475 = ??

Using the coefficients in the balanced equation, convert moles Al(OH)3 to moles H2SO4.
??moles Al(OH)3 x [3 moles H2SO4/2 moles Al(OH)3] = xx moles H2SO4.

Convert moles H2SO4 to grams H2SO4. g = moles x molar mass

[Note: Al(OH)3 is NOT a strong base, it isn't very soluble in water, and it is not corrosive. Neutralizing it with H2SO4 (a strong acid and VERY corrosive) is an extremely poor idea. I'm sure this was just a problem example but in practice it wouldn't be done.]

Answer 2

so moles Al(OH)3 = M x L = 0.500 x 0.0475 = ??

is 0.02375?

Answer 3

yes

Answer 4

okay so i got,

0.02375 molAl(OH)3x3molH2SO4/2molAl(OH)3=0.035625molH2SO4

0.035625molH2SO4x98.08g/molH2SO4= 3.4941g H2SO4 Correct? so the 3.4941g would be how much H2SO4 needed to neutralize the spill?

Answer 5

yes. However, since you are limited to 3 significant figures (from the 0.500 and 47.5 mL) I would round that to 3.49 g.

Answer 6

thanks!