How many bromine anions are there in 3.450 mol of MgBr2?
The correct answer and explanation is:
Let’s solve the problem step-by-step:
Question:
How many bromine anions (Br⁻) are there in 3.450 mol of magnesium bromide (MgBr₂)?
Step 1: Understand the compound formula
Magnesium bromide is MgBr₂, meaning each formula unit contains:
- 1 magnesium cation (Mg²⁺)
- 2 bromine anions (Br⁻)
Step 2: Determine the number of bromide ions per mole of MgBr₂
Each mole of MgBr₂ contains 2 moles of Br⁻ ions because of the subscript 2 in the formula.
Step 3: Calculate the moles of bromide ions
Given 3.450 mol of MgBr₂:
Moles of Br⁻ = 3.450 mol MgBr₂ × 2 mol Br⁻/1 mol MgBr₂
= 6.900 mol Br⁻
Step 4: Calculate the number of bromide ions
We use Avogadro’s number, which is approximately 6.022×10236.022 \times 10^{23} particles/mol, to convert moles to number of ions.
Number of Br⁻ ions = 6.900 mol × 6.022×10236.022 \times 10^{23} ions/mol
= 4.155×10244.155 \times 10^{24} bromide ions
Final answer:
There are approximately 4.16 × 10²⁴ bromide ions in 3.450 moles of MgBr₂.
Explanation:
This problem involves understanding the molecular formula and mole concept in chemistry. MgBr₂ dissociates into one magnesium ion and two bromide ions per formula unit. Since moles represent a count of particles (ions, atoms, or molecules), the number of bromide ions is double the number of moles of MgBr₂. By multiplying the given moles of MgBr₂ by 2, we get the moles of bromide ions. Then, using Avogadro’s number, we convert moles to actual number of ions. This approach is fundamental in stoichiometry, helping chemists quantify the exact number of particles involved in chemical reactions or compounds