Question

What ratio is used to carry out each conversion? a.mol CH4 to grams CH4 b.L CH4(g) to mol CH4(g)(at STP) c.molecules CH4 to mol CH4 I honestly don't kno what it's asking me I got 1 mol CH4/ 16(g) CH4 for a. 1 mol/22.4 L for b. and 6.02 x 10^23 molecules/1 mol but I don't think this is right, is it?

Answer 1

Step-by-step explanation:

According to mole concept

a.
`CH_4`Moles to
`CH_4`grams

Number of moles =
`\frac{\text{Mass of the compound}}{\text{Molar mass of the compound}}`

Mass of
`CH_4`= Moles of
`CH_4*` 16 g/mol

b.
`CH_4` Liters to
`CH_4` moles at STP.

At STP,1 mol gas occupies 22.4 L of volume.

Multiply the given volume of gas in L with
`(1 )/(22.4 ) moles`

c.Molecules
`CH_4`of to moles
`CH_4`

1 mole =
`N_A=6.022* 10^(23)` atoms or molecules

So, in 1 molecule =
`(1)/(N_A)=(1)/(6.022* 10^(23))` mole.

Multiply the given number of molecules with
`(1)/(6.022* 10^(23))` mole to get the number of moles of
`CH_4`.

Answer 2

a.mol CH4 to grams CH4
To convert from moles to grams, we need the molar mass of CH4 which is 16.05 g/mol.


b.L CH4(g) to mol CH4(g)(at STP)
To convert from L to mol CH4, we need the relation 1 mol of an ideal gas is equal to 22.4 L of that gas. The ratio would be 1/22.4 mol/L.


c.molecules CH4 to mol CH4
To convert from molecules to mol, we need the avogadro's number which is 6.022x10^23 units / 1 mol. The ratio to be multiplied would be 1/6.022x10^23 mol/molecules.