What mass of O2, in grams, is required for complete combustion of 25.5g of methane(CH4)?

CH4 + 2O2 --> CO2 + 2H2O

1) calculate the moles of CH4 in 25.5 g

2) you need twice as many moles of O2

calculate the mass of O2 in that many moles

3) The moles of CO2 = moles of CH4 and the moles of H2O = 2xthe moles of CH4

4) calculate the mass of CO2 and of H2O from the number of moles; sum the mass of CO2 and H2O

First write your balanced equation. This is a combustion equation. So when take methane (or anything when you combust it) and oxygen you will get carbon dioxide and water as products.

CH4 + 2 O2 --> CO2 + 2 H2O

Then find the MW of both CH4 and O2

C = 12.01 g/mol

H = 1.01 g/mol

O = 16.00 g/mol

So, MW for CH4 = 16.05 g/mol and O2 = 32.00 g/mol

Now start with what you are given, 25.5 g CH4 and convert to moles using the MW of CH4.

25.5 g CH4 x 1 mol CH4 / 16.05 g CH4 = 1.588 mol CH4

Now go from moles of CH4 to moles of O2 using your balanced equation

1.588 mol CH4 x 2 mol O2 / 1 mol CH4 = 3.177 mol O2

Now go from mole O2 to grams of O2 using the MW of O2

3.177 mol O2 x 32.00 g O2/ 1 mol O2 = 101.68 g O2

you should write the following equation:

CH4+2O2→CO2+2H2O

no. of mole of CH4=25.5/(12+4)=1.59375 mol

male ratio of :CH4:O2=1:2

no. of mole of O2=1.59375 X 2=3.1875 mol

MASS OF O2 REQUIRED= 16X2 X3.1875 =102 G

CH4 + O2 = CO2 + 2H2O

Considering molecular wt for each constituent,

(12+4) +2 (32 ) = (12+32 ) + 2(2+16 ).

16+ 64 = 44+ 36.

16/16 + 64/16 = 44/16 +36/16

So 1 kg CH4 Required 4 kg O2 AND product will be 2.75 kg CO2 + 2.25 KG Water vapor will be formed.

NOW, we have 25.5 gms CH4 = .0255 KG.

(1*0.0255 KG CH4) + (0.0255*4 KG O2 ) = (2.75* 0.0255 CO2 ) + ( 2.25 * 0.0255 WATER)

= 0.0255 KG CH4 + 0.102 KG O2 WILL PRODUCE 0.070125KG CO2 + 0.05737 WATER VAPOR.

= 25.5 GMS CH4 + 102 GMS O2 WILL PRODUCE 70.12 GMS CO2 + 57.37 GMS WATER VAPOR.