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

Answer:

No, there are not enough Ba atoms

Explanation:

Mass of $2.50cm^{3}$ of Ba = $(2.50\times 3.51)g=8.775g$

Mass of $1.75cm^{3}$ of S = $(1.75\times 2.07)g=3.6225g$

Molar mass of Ba = 137.33 g

Molar mass of S = 32.06 g

1 mol of an element contains $6.023\times 10^{23}$ number of atoms.

So 8.775 g of Ba = $\frac{8.775\times 6.023\times 10^{23}}{137.33}atoms=3.85\times 10^{22}atoms$

So 3.6225 g of S = $\frac{3.6225\times 6.023\times 10^{23}}{32.06}atoms=6.81\times 10^{22}atoms$

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