If 720 mL of 0.0165 M aqueous Al3+ and 300 mL of 0.0567 M aqueous Ca2+ are reacted, how many mol of Br- is produced?

Al2(SO4)3(aq) + 3 CaBr2(aq) → 2 AlBr3(aq) + 3 CaSO4(s)

my answer was .03564 mol of Br will be produced but the real answer is .0340 mol...i do not know what i did incorrectly

This looks like a limiting reactant question. How many moles of each reactant..

Al2(SO4)3 moles .720*.0165=.01188
CaBr2 .3*.0567=.01701
Hmmm. The balaced equation requires three times as much calcium bromide as aluminum sulfate.
So, the calcium bromide is the limiting reageant.

Moles of AlBr3 produced= 2/3 (.0170 and moles of Br- involved is 3*2/3*.0170 or .0340 moles.

I would like to point out a flaw in the question. If no matter how much Al^+3 reacts with no matter how much Ca^+2, then NO Br^- will be produced. You don't have any Br^- there to begin with.

To find the number of moles of Br- produced in the reaction, you need to use stoichiometry. Let's break down the steps to find the correct answer:

1. Calculate the number of moles of Al3+ and Ca2+:

Given:
- Volume of Al3+ solution: 720 mL
- Concentration of Al3+ solution: 0.0165 M

To find moles of Al3+:
moles of Al3+ = volume (in L) × concentration (in M)
moles of Al3+ = 720 mL × (1 L / 1000 mL) × 0.0165 M

2. Repeat the same calculation for Ca2+ solution:

Given:
- Volume of Ca2+ solution: 300 mL
- Concentration of Ca2+ solution: 0.0567 M

To find moles of Ca2+:
moles of Ca2+ = volume (in L) × concentration (in M)
moles of Ca2+ = 300 mL × (1 L / 1000 mL) × 0.0567 M

3. Determine the limiting reagent:

Now, we need to identify which reactant is the limiting reagent. The limiting reagent is the reactant that is completely consumed, thus limiting the amount of product formed.

Looking at the balanced equation:
Al2(SO4)3(aq) + 3 CaBr2(aq) → 2 AlBr3(aq) + 3 CaSO4(s)

The stoichiometric coefficient of Al2(SO4)3 is 1, and the stoichiometric coefficient of CaBr2 is 3. This means that 1 mole of Al2(SO4)3 reacts with 3 moles of CaBr2.

Calculate the number of moles of AlBr3 that can be formed from the given amount of Al3+ and Ca2+:
moles of AlBr3 = (moles of Al3+) × (2 / 1)

Calculate the number of moles of CaBr2 needed to react with that amount of AlBr3:
moles of CaBr2 needed = (moles of AlBr3) × (3 / 2)

Compare the moles of CaBr2 needed with the moles of CaBr2 given. The smallest value will indicate the limiting reagent.

4. Calculate the moles of Br- produced:

Now, calculate the moles of Br- produced from the limiting reagent, which is CaBr2:

moles of Br- produced = (moles of CaBr2) × (6 / 3)

Make sure to perform all calculations correctly to obtain the accurate answer.