posted by Nicole .
I don't know how to set up this problem...can you help?
How many milliliters of 6.00M HCl are required to react with 150 mL of 1.48 M Al(OH)3?
Al(OH)3+3HCl to AlCl3 + 3H2O
Each mole of Al(OH)₃ requires 3 moles of H. You have .15L * 1.48M Al(OH)₃ = 0.222mol of Al.
So, you need 3*0.222 = 0.666mol of H to react with it.
1L of 6M HCl has 6mol of H. You need 0.666 mol, so that would be .666/6 = 0.111L or 111ml of HCl
... I think.
111 mL is correct.