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Tanzania [10]
3 years ago
11

How many MOLES of boron tribromide are present in 3.20 grams of this compound ?

Chemistry
1 answer:
Dmitriy789 [7]3 years ago
7 0

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

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