Question

Question #3063a

Answer 1

The percentages by mass of `CO_2` and `Ne` are 91.6 and 8.4 respectively.

The number of moles of a substance is its mass m divided by the mass of 1 mole which is its Mr in grams.

So no. moles `n = m/M_r`

`M_r[CO_2]=44` `A_r[Ne]=20`

And

`m_(CO_2) + m_(Ne)= 66.0g`

Lets call this equation (1)

So we can write the total number of moles of the mixture as:

Total moles = `m_(CO_2)/(44)+m_(Ne)/(20)=1.65`

Lets call this equation(2)

So now we have a situation where we have 2 equations and 2 unknowns. These are called simultaneous equations.

From equation (1) we can write:

`m_(CO_2)=66- m_(Ne)`

We can now substitute this expression for `m_(CO_2` into equation (2):

`rArr` `((66- m_(Ne)))/(44)+m_(Ne)/(20)=1.65`

So `66/44-m_(Ne)/(44)+m_(Ne)/20=1.65`

So `1.5-0.023m_(Ne)+0.05m_(Ne)=1.65`

From which `m_(Ne)=5.55g`

So `m_(CO_2) =66.0-5.55=60.45g`

So % `CO_2`by mass `=(60.45)/66.0`x 100 = 91.6%

So % `Ne` by mass = (100-91.6) = 8.4%