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If 2.8 moles of CH4 react with 3 moles of O2 what is the limiting reagent?
Wiki User
∙ 9y ago
Take the balanced equation. CH4+2O2---->CO2+2H2O.So 2.8mol of CH4 need 5.6mol of O2.So O2 is limitting factor
Wiki User
∙ 9y ago
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What is the limiting reactant in combustion of methane is 2.8 moles of methane and 5 moles of oxygen?
Methane reacts with oxygen in the following way. CH4 + 3 O2 --> CO2 + 4 H2O. If 5 moles of oxygen react with 2.8 moles of methane, only 1.67 moles of methane would be consumed because of the molar ratio 1:3. This would produce 1.67 moles of carbon dioxide and 6.67 moles of water.
How many moles of co² produced by the reaction of 2.0 moles of CH⁴ with excess O²?
Balanced equation. CH4 + 2O2 -> CO2 + 2H2O one to one, but set-up anyway and limiting reacting known 2.0 moles CH4 (1 mole CO2/1 mole CH4) = 2.0 moles CO2 produced ( by the way, those numbers are subscripts, not superscripts )
How many molecules of oxygen 2 are needed to react with which molecule of methane CH4?
6 molecules of oxygen are needed to react with 3 methane molecules as one molecule of oxygen ( O2) are needed for methane gas.
How many kcal are produced when 32.0 g of CH4 react?
How many kcal are produced when 32.0g of CH4 react? CH4+2O2->CO2+2H2O+218 kcal
Are there two moles in CH4?
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.
What is the limiting reactant in combustion of methane is 2.8 moles of methane and 5 moles of oxygen?
Methane reacts with oxygen in the following way. CH4 + 3 O2 --> CO2 + 4 H2O. If 5 moles of oxygen react with 2.8 moles of methane, only 1.67 moles of methane would be consumed because of the molar ratio 1:3. This would produce 1.67 moles of carbon dioxide and 6.67 moles of water.
When 3.05 of ch4 is mixed with 5.03 moles of o2 the limiting reactants is?
cheaters will not succeed
How many moles of co² produced by the reaction of 2.0 moles of CH⁴ with excess O²?
Balanced equation. CH4 + 2O2 -> CO2 + 2H2O one to one, but set-up anyway and limiting reacting known 2.0 moles CH4 (1 mole CO2/1 mole CH4) = 2.0 moles CO2 produced ( by the way, those numbers are subscripts, not superscripts )
How many moles of product would be produced if 0.500 mole of the first reactant were to react completely CO2 4H2 CH4 2H2O?
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====================
How many moles of CH4 are in 200 grams of CH4?
200 g CH4 x 1 mole CH4/16 g = 12.5 moles CH4
How many moles of hydrogen gas would be needed to react with excess carbon dioxide to produce 53.6 moles of water vapor?
This is the Sabatier process and is usually in need of high temperature and a metallic catalyst. So, balanced equation. CO2 + 4H2 -> CH4 + 2H2O As said, hydrogen is in excess, so CO2 is limiting and drives the reaction. One to one 36.6 moles CO2 (1 mole CH4/1 mole CO2) = 36.6 moles CH4 produced in this reaction -------------------------------------------------------
How many molecules of oxygen 2 are needed to react with which molecule of methane CH4?
6 molecules of oxygen are needed to react with 3 methane molecules as one molecule of oxygen ( O2) are needed for methane gas.
When 85.0 g of CH4 are mixed with 160 g of O2 the limiting reactant is?
When 85.0 g of CH4 are mixed with 160. g of O2 the limiting reactant is __________. CH4 + 2O2 → CO2 + 2H2O
How many grams of CO2 are formed for every gram of CH4 that combusts?
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.
How many moles in 64 grams of CH4?
There are 0.75 moles in it.You have to devide 12 by molecular mass
24g of CH4 react with 96 g of O2?
CH4 + 2O2 gives CO2 and 2H2O.So 16g of CH4 react with 64g of O2.So 24g react with 96g of O2
How many atoms of hydrogens are in 3 moles of CH4?
In 3 moles of CH4, there are 18.06 x 10^23 times Hydrogen atoms.
