Question #8a010

2 Answers
Jul 21, 2015

Your reaction will produce 100 g of carbon dioxide.

Explanation:

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

CH_(4(g)) + 2O_(2(g)) -> CO_(2(g)) + 2H_2O_((l))CH4(g)+2O2(g)CO2(g)+2H2O(l)

Notice that you have a 1:11:1 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)

Jul 21, 2015

110g

Explanation:

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)