Question #3063a

Question

1881 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-10-13T22:31:30

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%