Answer:
276g of Br are present
Explanation:
To solve this question we need to find the moles of CaBr2 using its molar mass -CaBr2: 199.89g/mol-. As 1 mole of CaBr2 contains 2 moles of Br we can find the moles of Br and its mass:
Moles CaBr2:
345g * (1mol/199.89g) = 1.7259 moles CaBr2
Moles Br:
1.7259 moles CaBr2 * (2mol Br / 1mol CaBr2) = 3.4519 moles Br
Mass Br -Molar mass: 79.904g/mol-
3.4519 moles Br * (79.904g/mol) = 276g of Br are present
