the lead (II) nitrate in 25.49 ml of a 0.1338M solution reacts with all of the aluminum sulfate in 25.00 ml solution. What is the molar concentration of the aluminum sulfate in the original aluminum sulfate solution?

The reaction is a 3 to 1 ratio Pb(NO3)2 : Al2(SO4)3 => moles aluminum sulfate = 1/3(moles lead nitrate) => 1/3[(0.02549L)(0.1338M)] = 0.00114 mole aluminum sulfate => Therefore, concentration = [Al2(SO4)3] = 0.00114 mole/0.02549L = 0.045M Aluminum Sulfate.

To find the molar concentration of the aluminum sulfate in the original solution, you can use the concept of stoichiometry, which relates the amount of reactants and products in a chemical reaction.

First, let's write down the balanced equation for the reaction between lead (II) nitrate (Pb(NO3)2) and aluminum sulfate (Al2(SO4)3):

3Pb(NO3)2 + 2Al2(SO4)3 -> 3PbSO4 + 2Al(NO3)3

From the balanced equation, we can see that the ratio between lead (II) nitrate and aluminum sulfate is 3:2. This means that for every 3 moles of lead (II) nitrate that react, 2 moles of aluminum sulfate will react.

Next, we need to determine the number of moles of lead (II) nitrate in the 25.49 ml of 0.1338M solution. To do this, multiply the volume (in liters) by the molar concentration:

moles of Pb(NO3)2 = volume (in liters) * molar concentration
= 0.02549 L * 0.1338 mol/L
= 0.003414 mol

Since the reaction is stoichiometrically balanced, we know that the molar ratio of lead (II) nitrate to aluminum sulfate is 3:2. Therefore, the number of moles of aluminum sulfate must be (2/3) times the number of moles of lead (II) nitrate:

moles of Al2(SO4)3 = (2/3) * moles of Pb(NO3)2
= (2/3) * 0.003414 mol
= 0.002276 mol

Finally, we need to calculate the molar concentration of the aluminum sulfate in the original solution. We know that the final volume is 25.00 ml, so:

molar concentration of Al2(SO4)3 = moles of Al2(SO4)3 / volume (in liters)
= 0.002276 mol / 0.025 L
= 0.09104 M

Therefore, the molar concentration of the aluminum sulfate in the original aluminum sulfate solution is 0.09104 M.