Question

You need 1.50 moles of ammonia (NH3) for a reaction at STP. What volume of ammonia must you measure?

Also,

How many molecules of NaCl are in 23.40 grams NaCl?

Answer 1

I have an idea on how to work it out, but I'll be honest I'm not that good at working these out. But have you taken a look at Bitesize? It's a really good website and should help you. Just search 'how to work out moles in chemistry' or 'moles chemistry bitesize' something like that. Hope this helps, if not come back to me :)

Answer 2

1) 33.64 Liter will be volume of ammonia.

2)
`2.409* 10^(23)` molecules of NaCl are in 23.40 grams NaCl.

Step-by-step explanation:

1) Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas =
`1 atm` (at STP)

V = Volume of gas = ?

n = number of moles of gas = 1.50 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 273,15 K

Putting values in above equation, we get:

`V=(nRT)/(P)=(1.50 mol* 0.0821 atm L/mol K* 273.15 K)/(1 atm)`

V = 33.64 L

2)
`N=n* N_A`

Where:

N = Number of particles / atoms/ molecules

n = Number of moles

`N_A=6.022* 10^(23) mol^(-1)` = Avogadro's number

We have:

Molar mass of NaCl = 58.5 g/mol

n =
`(23.40 g)/(58.5 g/mol)=0.4 mol`

`N=0.4 mol* 6.022* 10^(23) mol^(-1)=2.409* 10^(23) molecules`

`2.409* 10^(23)` molecules of NaCl are in 23.40 grams NaCl.