Question

A chemist determines by measurements that 0.0800 moles of bromine liquid participate in a chemical reaction. Calculate the mass of bromine liquid that participates. Round your answer to 3 significant digits.

Answer 1

Answer:

The mass of bromine liquid that participates in  a chemical reaction=12.8 g

Explanation:

We are given that

Total number of moles of bromine liquid participate in chemical reaction=0.0800 moles

We have to find the mass of bromine liquid that participates.

Atomic mass of Br=79.9 g

1  mole of bromine liquid=2 atomic mass of bromine (Br)

1 mole of bromine liquid ($Br_2$) =$2\times 79.9=159.8 g$

0.0800 moles of  bromine liquid=$159.8\times 0.0800$ g

0.0800 moles of  bromine liquid=12.784 g

0.0800 moles of  bromine liquid$\approx 12.8$ g

Hence, the mass of bromine liquid that participates in  a chemical reaction=12.8 g