A solution of 0.20 mole of NaBr and 0.20 mole of MgBr 2 in 2.0
A solution of 0.20 mole of NaBr and 0.20 mole of MgBr2 in 2.0 dm3 of water is to be analysed. How many moles of Pb(NO3)2 must be added to precipitate all the bromide as insoluble PbBr2?
- A) 0.30 mol
- B) 0.10 mol
- C) 0.20 mol
- D) 0.40 mol
Correct Answer: A) 0.30 mol
Explanation
In this question, we have a solution containing 0.20 moles of NaBr and 0.20 moles of MgBr\(_2\) in 2.0 dm\(^3\) of water. We are asked to find out how many moles of Pb(NO\(_3\))\(_2\) must be added in order to precipitate all the bromide as insoluble PbBr\(_2\). To solve this problem, we need to use the concept of stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction. In this case, we know that Pb(NO\(_3\))\(_2\) reacts with NaBr and MgBr\(_2\) to form PbBr\(_2\), which is insoluble and precipitates out of the solution. The balanced chemical equation for this reaction is: Pb(NO\(_3\))\(_2\) + 2NaBr + MgBr\(_2\) ? PbBr\(_2\) + 2NaNO\(_3\) + Mg(NO\(_3\))\(_2\) From this equation, we can see that 1 mole of Pb(NO\(_3\))\(_2\) reacts with 2 moles of NaBr and 1 mole of MgBr\(_2\) to produce 1 mole of PbBr\(_2\). Therefore, to precipitate all the bromide as PbBr\(_2\), we need to add an amount of Pb(NO\(_3\))\(_2\) that is equal to the total number of moles of NaBr and MgBr\(_2\) in the solution. The total number of moles of NaBr and MgBr\(_2\) in the solution is: 0.20 mol (NaBr) + 0.20 mol (MgBr\(_2\)) = 0.40 mol Therefore, we need to add 0.40 moles of Pb(NO\(_3\))\(_2\) to precipitate all the bromide as PbBr\(_2\). The correct answer is option A: 0.30 mol. This answer is incorrect given the explanation. The correct answer should be option D: 0.40 mol.