Question

How many aluminum ions are there in 1.46 mol Al2S3? Express your answer to three significant figures.

Answer 1

Answer: There are
`17.6* 10^(23)` alumnium ions in 1.46 mol
`Al_2S_3`

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
`6.023* 10^(23)` of particles.

1 mole of
`Al_2S_3` contains = 2 moles of
`Al^(3+)` ions

Thus 1.46 mole of
`Al_2S_3` contains =
`(2)/(1)* 1.46=2.92` moles of
`Al^(3+)` ions

Now 1 mole of
`Al^(3+)` contains =
`6.023* 10^(23)`
`Al^(3+)` ions

Thus 2.92 moles of
`Al^(3+)` contains =
`(6.023* 10^(23))/(1)* 2.92=17.6* 10^(23)`
`Al^(3+)` ions.

There are
`17.6* 10^(23)` alumnium ions in 1.46 mol
`Al_2S_3`