Answer:
1) There are 0.0128 moles of BBr3 present in 3.20 grams of this compound
2) There are 909.35 grams of BBr3 present in 3.63 moles of this compound
Explanation:
<u>Step 1: </u>Given data
Boron tribromide = BBr3
Molar mass of Boron = 10.81 g/mole
Molar mass of Bromide = 79.9 g/mole
Molar mass of Boron tribromide = 10.81 + 3*79.9 = 250.51 g/mole
<u>Step 2:</u> Calculating number of moles
Number of moles = mass / molar mass
Number of moles of BBr3 = 3.20 grams / 250.51 g/mole
Number of moles of BBr3 = 0.0128 moles
<u>Step 3:</u> Calculating mass
If we have 3.63 moles of boron tribromide (= BBr3)
Mass of BBr3 = number of moles of BBr3 * Molar mass of BBr3
Mass of BBr3 = 3.63 moles * 250.51 g/mole
Mass of BBr3 = 909.35 grams
