Answer:
3.39g of MgBr
Explanation:
Reaction between magnesium and bromine
Mg + Br₂ → MgBr
Mass of MgBr = ?
1g of Mg + 5.0g of Br₂
6g of Mg and Br₂
Molar mass of Mg = 24g/mol
Molar mass of Br = 80g/mol
Number of moles = mass / molar mass
Mass = number of moles * molarmass
Mass of Mg = 1 * 24 = 24g
Mass of Br₂ = 1 * (2*80) = 160g (diatomic molecule)
Molarmass of MgBr = (24 + 80) = 104g/mol
Mass of MgBr = 1 * 104 = 104g/mol
24g of Mg + 160g of Br₂ = 104g of MgBr
184g of (mg and Br) = 104g of mgBr
6g of (mg and Br) = y g of MgBr
y = (6 * 104) / 184
y = 3.39g of MgBr
