1. Find the molar mass of BaBr₂:
* Barium (Ba) has a molar mass of 137.33 g/mol
* Bromine (Br) has a molar mass of 79.90 g/mol
* Since there are two bromine atoms, the total mass of bromine is 2 * 79.90 g/mol = 159.80 g/mol
* The molar mass of BaBr₂ is 137.33 g/mol + 159.80 g/mol = 297.13 g/mol
2. Use the molar mass to convert moles to grams:
* Grams = Moles * Molar Mass
* Grams = 3.56 moles * 297.13 g/mol
* Grams ≈ 1058.6 g
Therefore, approximately 1058.6 grams of barium bromide are present in 3.56 moles of the compound.
