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?
For a question like this how do I find out it is a 3:1 ratio without the balance equation?
To determine the ratio between lead (II) nitrate and aluminum sulfate, you need to use the balanced chemical equation for the reaction. Without the balanced equation, it is not possible to determine the exact ratio between the two compounds.
However, if you know that the reaction between lead (II) nitrate and aluminum sulfate is a 1:1 stoichiometric ratio, meaning one mole of lead (II) nitrate reacts with one mole of aluminum sulfate, you can use this information to determine the molar concentration of aluminum sulfate in the original solution.
Given that the lead (II) nitrate solution has a concentration of 0.1338 M and a volume of 25.49 ml, and assuming a 1:1 stoichiometric ratio, you can calculate the amount of lead (II) nitrate used in the reaction:
Amount of lead (II) nitrate = Concentration x Volume
= 0.1338 M × 0.02549 L
= 0.003413662 mole
Since the reaction is a 1:1 stoichiometric ratio, the amount of aluminum sulfate used in the reaction will also be 0.003413662 mole.
Now, to find the molar concentration of aluminum sulfate in the original solution, you need to calculate its concentration using its volume. Given that the aluminum sulfate solution has a volume of 25.00 ml:
Concentration of aluminum sulfate = Amount of aluminum sulfate / Volume
= 0.003413662 mole / 0.025 L
= 0.13654648 M
Therefore, the molar concentration of aluminum sulfate in the original aluminum sulfate solution is approximately 0.1365 M.
To determine the molar concentration of aluminum sulfate in the original solution, we need to use the principle of stoichiometry and the concept of balanced chemical equations.
In this case, we're given that 25.49 mL of a 0.1338 M lead (II) nitrate solution reacts with all of the aluminum sulfate in a 25.00 mL solution. The key to finding the concentration of aluminum sulfate is to use the stoichiometric ratio between the reactants.
However, without the balanced equation, it is not possible to directly determine the stoichiometric ratio. Therefore, we first need to find the balanced equation for the reaction.
Let's assume that the balanced equation for the reaction between lead (II) nitrate (Pb(NO3)2) and aluminum sulfate (Al2(SO4)3) is:
3Pb(NO3)2 + 2Al2(SO4)3 → 3PbSO4 + Al2(NO3)6
This balanced equation shows that 3 moles of lead (II) nitrate react with 2 moles of aluminum sulfate. From this information, we can establish the stoichiometric ratio as 3:2 between lead (II) nitrate and aluminum sulfate.
Now, using the given information, we can set up a ratio of concentrations to find the molar concentration of aluminum sulfate:
(0.1338 M lead (II) nitrate) / (25.49 mL) = (x M aluminum sulfate) / (25.00 mL)
Simplifying this ratio, we get:
(0.1338 M / 25.49 mL) = (x M / 25.00 mL)
Next, we can cross multiply and solve for x:
x = (0.1338 M / 25.49 mL) * 25.00 mL
x ≈ 0.13180 M
Therefore, the molar concentration of the aluminum sulfate in the original aluminum sulfate solution is approximately 0.13180 M.