The amount of oxygen is 0,067 moles.
In STP a mole has a volume of 22.4dm3.so 1.5 mole has a volume of 33.6dm3
33.6 liters
200 g CH4 x 1 mole CH4/16 g = 12.5 moles CH4
CH4 + 2 H2O = 3 H2 + CO2 8 moles CH4 produce 8 x 3 moles H2, which is 24.
CO2 + 4H2 --> CH4 + 2H2O0.500 moles CO2 (1 mole CH4/1 mole CO2) = 0.500 moles CH40.500 moles CO2 (2 moles H2O/1 mole CO2) = 1.00 moles H2O-------------------------------------------------------------------------------------add= 1.50 moles total product====================
2,8 moles is of course equivalent to 2,8 moles !Probable is a spelling error in your question.
156 grams CH4 (1 mole CH4/16.042 grams) = 9.72 moles of methane ==================
200 g CH4 x 1 mole CH4/16 g = 12.5 moles CH4
The balanced equation for combustion of CH4 is CH4 + 2O2 ==> CO2 + 2H2OThus, one mole CH4 produces 1 mole CO21 g CH4 x 1 mole CH4/16 g = 0.0625 moles CH40.0625 moles CH4 ==> 0.0625 moles CO20.0625 moles CO2 x 44 g CO2/mole = 2.75 g CO2Thus, the answer would be that 1 grams of CH4 will produce 2.75 grams of CO2 after complete combustion.
There are 0.75 moles in it.You have to devide 12 by molecular mass
In 3 moles of CH4, there are 18.06 x 10^23 times Hydrogen atoms.
CH4 + 2 H2O = 3 H2 + CO2 8 moles CH4 produce 8 x 3 moles H2, which is 24.
CO2 + 4H2 --> CH4 + 2H2O0.500 moles CO2 (1 mole CH4/1 mole CO2) = 0.500 moles CH40.500 moles CO2 (2 moles H2O/1 mole CO2) = 1.00 moles H2O-------------------------------------------------------------------------------------add= 1.50 moles total product====================
That's a tricky question, because one molecule of CH4 is simply that, one atom of carbon and 4 atoms of Hydrogen. Moles are a UNIT used to transform atoms (which we cannot measure individually in the lab) into practical units such as grams (which we can measure). The moles of CH4 depend on the mass, in SI units of grams, that you have of this substance. The molecular weight of CH4 is 16 g/mol (12 for Carbon + 1 for each Hydrogen). If you WANTED 2 moles of CH4, you need to multiply this molecular weight by 2 moles to get 32 grams (the moles cancel out upon multiplication). So, 32 grams of CH4 is 2 moles of CH4.
Take the balanced equation. CH4+2O2---->CO2+2H2O.So 2.8mol of CH4 need 5.6mol of O2.So O2 is limitting factor
2,8 moles is of course equivalent to 2,8 moles !Probable is a spelling error in your question.
156 grams CH4 (1 mole CH4/16.042 grams) = 9.72 moles of methane ==================
Divide 96 by molecular mass.So the answer is 6mol
4.66715966 × 1024