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.

To set up this problem, we need to use the concept of stoichiometry. Stoichiometry allows us to relate the amounts of substances involved in a chemical reaction.

First, let's write down the balanced equation for the reaction:

Al(OH)3 + 3HCl → AlCl3 + 3H2O

From the equation, we can see that 1 mole of Al(OH)3 reacts with 3 moles of HCl. Therefore, we need to determine the moles of Al(OH)3 and convert it into the moles of HCl using this ratio.

To find the moles of Al(OH)3, we use the formula:
moles = concentration (in M) × volume (in liters)

Given that the concentration of Al(OH)3 is 1.48 M and the volume is 150 mL, convert the volume to liters:
150 mL = 150/1000 L = 0.150 L

Now, calculate the moles of Al(OH)3:
moles of Al(OH)3 = 1.48 M × 0.150 L = 0.222 moles

Since the Al(OH)3 to HCl ratio is 1:3, we can see that we will need 3 times the moles of HCl as compared to Al(OH)3.

Finally, we can calculate the volume of 6.00 M HCl required. We will use the same formula as before, but this time solving for the volume:
volume (in liters) = moles/concentration (in M)

Given that the concentration of HCl is 6.00 M, substitute the values:
volume (in liters) = 0.222 moles / 6.00 M = 0.037 L

The volume is currently in liters, so let's convert it to milliliters:
0.037 L = 0.037 × 1000 mL = 37 mL

Therefore, 37 milliliters of 6.00 M HCl are required to react with 150 mL of 1.48 M Al(OH)3.