If 80.0g of CH4 is consumed by the reaction CH4 +2O2 = CO2 +2H2O, what mass of CO2 is produced?

Question

Use 12.01 amu for the atomic mass of C, 1.008 amu for H and 16 amu for O.
I know that the answer is 219g, i just dont know how to get it?

Answer 1

80g/(12.01+4*1.008) = x/(12.01+2*16)
80g/16.042 = x/44.01
x = 44.01*80.0/16.042
x = 219.47 g
x ≈ 219 g

Answer 2

The molecular weight of CH4 is : 12.01g (C) + 4*1.008g (H) = 16.042g/mol

This means: in 80.0g of CH4, there are approximately 5 mols (precisely, 4.986) of CH4.

This ALSO means, there are 5 mols of carbon as well.

If you look at the right side of the reaction, the only place carbon pops up is for the carbon dioxide.

Therefore, the 5 moles of carbon that is used in the reaction all goes to the carbon dioxide.

Per one mole of carbon dioxide, there is :12.01 (C) + 16*2 (O) = 44.01 g/mol.

Multiply: 44.01g/mol * 4.9869 mols = 219.4 g/mol

I hope the explanation makes sense.

Answer 3

Look
1 mole of ch4=16g
Therefore 80g=5 moles

Now if oxygen is present in unlimited qty then ch4 becomes limiting reactant,hence 5moles of co2 is formed.
1 mole of co2=44g
5mole =220g

there u go.
Give me ten for that

Answer 4

CH4 + 2O2 --> CO2 +2H2O
the molar ratio is 1:2:1:2
80.0g*mol/16.042g= 4.987mol
therefore, 4.987 mol of CO2 is produced
4.987mol*44.01g/mol=219.47787g

Answer 5

simple cross multiplication:

1 mole of CH4 (12.01 + [1.008x4]) = 16.042
1 mole of CO2 (12.01 + [16x2]) = 44.01

CH4 +2O2 ------------------> CO2 +2H2O

if16.042 gm of CH4 produce 44.01 gm of CO2
then 80 gm oc CH4 produce X gm of CO2

X = (44.01 x 80) / 16.042
X = 219.4

glad to help, if u need anything else just mail me..

Answer 6

left side

c=1
h=4
0=4

right side

c=1
H=4
0=4

they are equal so you do a straight maths equation

80= (12+4)x
x=5 so 5 molecules of CH4 were used
so everything is timesed by 4
hmm i got 176g for C02 and 248g for the total mass of products formed

i dunno

Answer 7

same amount because of same # of mlcls.