Question

When barium (Ba) reacts with sulfur (S) to form barium sulfide (BaS), each Ba atom reacts with an 5 atom. If 250 cm3 of Ba reacts with L75 cm3 of S. are there enough Ba atoms to react with the S atoms (d of Ba = 3.51 g/cm3; dof S = 207 g/cm3)?

Answer 1

No, there are not enough Ba atoms

Step-by-step explanation:

Mass of
`2.50cm^(3)` of Ba =
`(2.50* 3.51)g=8.775g`

Mass of
`1.75cm^(3)` of S =
`(1.75* 2.07)g=3.6225g`

Molar mass of Ba = 137.33 g

Molar mass of S = 32.06 g

1 mol of an element contains
`6.023* 10^(23)` number of atoms.

So 8.775 g of Ba =
`(8.775* 6.023* 10^(23))/(137.33)atoms=3.85* 10^(22)atoms`

So 3.6225 g of S =
`(3.6225* 6.023* 10^(23))/(32.06)atoms=6.81* 10^(22)atoms`

As 1 atom of Ba reacts with 1 atom of S therefore enough Ba atoms are not present for reaction.