Question #8a010

Start by writing the balanced chemical equation for the combusiont of methane

`CH_(4(g)) + 2O_(2(g)) -> CO_(2(g)) + 2H_2O_((l))`

Notice that you have a `1:1` mole ratio between methane and carbon dioxide. This means that the number of moles of carbon dioxide produced will be equal to the number of moles of methane that reacted.

To determine how many moles of methane react, use the compound's molar mass

`40cancel("g") * ("1 mole "CH_4)/(16.04cancel("g")) = "2.49 moles"` `CH_4`

The reaction will thus produce

`2.49cancel("moles"CH_4) * ("1 mole "CO_2)/(1cancel("mole"CH_4)) = "2.49 moles"` `CO_2`

To get the mass of carbon dioxide that would contain this many moles, use the compound's molar mass

`2.49cancel("moles") * "44.01 g"/(1cancel("mole")) = "109.6 g"` `CO_2`

Rounded to one sig fig, the number of sig figs you gave for the mass of methane, the naswer will be

`m_(CO_2) = color(green)("100 g "CO_2)`

First we need an equation...

Methane + oxygen `rarr` carbon dioxide + water

`CH_(4(g)) + 2O_(2(g))rarrCO_(2(g)) + 2H_2O_((g))`

Work out the Mr of each...

  16.               64.          44.             16

So we can now say that if 16g of methane is burnt 44g of carbon dioxide will be produced. If we had 40g methane how much carbon dioxide would be produced?

From then on its a ratios question...

16:44

40:x

To go from 40 to 16 you divide by 2.5 (40/16=2.5) so to go from 44 to x you need to times by the same number (44x2.5=110)