Question

if 25.26 mL of a 0.125 M solution of aluminum hydroxide reacts exactly with 22.00 mL of a sulfuric acid solution, what is the molarity of the sulfuric acid solution?

3H2SO4(aq) + 2Al(OH)3(aq) → Al2(SO4)3(aq) + 6H20(1) a) How many grams of aluminum hydroxide will react with 75,0 mL of 0.319 M sulfuric acid solution? moles of H₂SO4 0.3198 0.075 0.0239 moles 3 moles of H2Soy requires 2 moles Al(OH)3 0.023a moles of the 504 - 2xo.0239 moles of Al(OH)₃ = 0.01595x78 = 11.2449 of Al 10Hz / b) If 25.26 mL of a 0.125 M solution of aluminum hydroxide reacts exactly with 22.00 mL of a sulfuric acid solution, what is the molarity of the sulfuric acid solution?

Answer 1

(a) To find the grams of aluminium hydroxide (Al(OH)) Given that 75.0mL of 0.319M sulfuric acid(H,SO.) solution 3H,50. (aq) +2A1(OH), (aq) → A1, (50.), (aq)+6H,0(1) First find the moles of H2SO solution moles = molarity x volume 75.0mL = 0.075L moles = 0.319mol/L 0.075L = 0.0239mol find the moles of Al(OH) According to reaction Al(OH), and H2S0, is 2:3 mole ratio 2 moles of Al(OH) Therefore moles of Al(OH), = 0.0239 mol of H2SO, 3 = 0.01593moles of Al(OH) 3 moles of H2SO4 To find the grams of Al(OH), mass = moles x molarmass molarmass of Al(OH) = 78g/mol mass = 0.01593mol x 78g/mol = 1.24g Hence the grams of Al(OH), = 1.24g|

(b) To find the molarity of sulfuric acid(H,50.) solution Given that 22.00 mL of sulfuric acid(H,50.) solution 25.26 mL of 0.125Ma solution of aluminium hydroxide (Al(OH)) 3H,90 (aq) + 2Al(OH), (aq) → A1, (SO4)2 (aq) +6H,0(1) First find them oles of aluminium hydroxide (Al(OH)) 25.26 mL = 0.02526L moles = molarity x volume = 0.125mol/L 0.02526L = 0.003157moles Findthe moles of H2SO4 According to reaction H,SO, and Al(OH), is 3.2 mole ratio 3 moles of H,SO Therefore moles of H2SO4 = 0.003157m ol of Al(OH), X- -= 0.004735m oles of H2SO4 2 moles of Al(OH), To find the molarity of sulfuric acid(H,50.) solution moles of solute Molarity = liter of solution 22.00 mL = 0.022L 0.004735moles Molarity= 0.022L = 0.215M Hence the molarity of sulfuric acid (H2SO4) solution = 0.215M