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.
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.]
so moles Al(OH)3 = M x L = 0.500 x 0.0475 = ??
is 0.02375?
yes
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?
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.
thanks!
To determine the amount of grams of H2SO4 needed to neutralize the spill of Al(OH)3, we need to use stoichiometry and the balanced chemical equation for the reaction between Al(OH)3 and H2SO4, which is:
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
From the balanced equation, we can see that it takes 3 moles of H2SO4 to react with 2 moles of Al(OH)3.
First, let's calculate the moles of Al(OH)3 in the spill:
Molarity (M) = mol/L
mol = M × L
mol = 0.500 mol/L × 0.0475 L = 0.02375 mol
Now, using the mole ratio from the balanced equation, we can determine the moles of H2SO4 needed:
Moles of H2SO4 = (moles of Al(OH)3) × (3 moles of H2SO4 / 2 moles of Al(OH)3)
Moles of H2SO4 = 0.02375 mol × 3/2 = 0.035625 mol
Finally, to calculate the grams of H2SO4 needed, we can use the molar mass of H2SO4:
Molar mass of H2SO4 = 2(1.008 g/mol of H) + 32.06 g/mol of S + 4(16.00 g/mol of O) = 98.09 g/mol
Grams of H2SO4 = (moles of H2SO4) × (molar mass of H2SO4)
Grams of H2SO4 = 0.035625 mol × 98.09 g/mol ≈ 3.50 grams
Therefore, approximately 3.50 grams of H2SO4 would be needed to neutralize the spill of Al(OH)3.