How many bromide ions (Br^-) are present in 3.50 g of MgBr_2?

• A. 0.037 mol
• B. 0.047 mol
• C. 0.074 mol
• D. 0.019 mol

Answer: (A)

Verified step by step guidance

1. Calculate the molar mass of magnesium bromide (MgBr\_2) by adding the atomic masses of magnesium (Mg) and two bromine (Br) atoms. Use the periodic table values: Mg ≈ 24.31 g/mol and Br ≈ 79.90 g/mol. So, molar mass = 24.31 + 2 \times 79.90.
2. Determine the number of moles of MgBr\_2 in 3.50 g by using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). Substitute the mass (3.50 g) and the molar mass calculated in step 1.
3. Recognize that each formula unit of MgBr\_2 contains 2 bromide ions (Br\^-). Therefore, the number of moles of bromide ions is twice the number of moles of MgBr\_2.
4. Calculate the moles of bromide ions by multiplying the moles of MgBr\_2 by 2.
5. Interpret the result as the amount of bromide ions in moles present in 3.50 g of MgBr\_2, which can be compared to the given answer choices.