What is the reaction between methane and chlorine in the presence of sunlight?
CH4 + CL2 = CH3Cl(Mono-Chloro-Methane) + HCl.The above reaction is an example of Chlorination. Chlorination is the substitution of hydrogen atoms by chlorine atom.The conditions used in the above process is U.V.light or diffused sunlight or 400℃.The reactions if the reaction is continued :CH3Cl + Cl2 = CH2Cl2 + HCl.CH2CL2 + Cl2 = CHCl3 + HCl.CHCl3 + Cl2 = CCl4 + HCl.The conditions for the reaction is same as the one in the start. By these reaction we get the product as covalent compound Carbon-tetra chloride.
1 mole of hydrocarbon burns completely in oxygen to produce 2 moles of CO2 and 2 mols of H2O. What is the formula for the hydrocarbon?
It is given 1 mole of the hydrocarbon gives 2 moles of CO2 and 2 moles of H2O on complete combustion.From this it is clear that 1 molecule of the hydrocarbon gives 2 molecules of CO2 (2CO2) and 2 molecules of H2O (2H2O) as products on complete combustion.It is evident from the above data that 1 molecule of the hydrocarbon is composed of 2 carbon and 4 hydrogen atoms.So, the hydrocarbon must have the molecular formula C2H4. It is ethene or ethylene, the simplest alkene.Equation for combustion: C2H4 + 3O2 = 2CO2 + 2H2O
What volume of oxygen is required for the complete combustion of a mixture of 50cm3 of CH4 and 50cm3 of C2H4?
Assuming STP though it really doesn't matter since gases have volumes relative to their molar weight.Getting your head around it.At STP we have.1 mole =22,400 cm3. (22.4 liters)Of course all gases are affected by temperature or pressure and change equally.But at STP we have for the methane or ethylene for that matter50 cm3/22,400 cm3 = 1/448 moles= 0.002232142857143 molesOf course this value changes with temperature and pressure. Since you were asked for volume that doesn't matter. It doesn't matter because when you measure volumes of gases they all change together and the molar ratios don't change.Methane molar massCH4 → 12 + 4 * 1 = 16At STP the mass would be.16/448= 1/28 grams = 0.035714285714286 gramsDeal only with volumes.Since you weren't given temperature or pressure or mass so you don't know precisely how much.Regardless you know the precise ratios!! Which is the point.The reactionsMethaneC + O2 → CO212 + 2 * 16 = 44 grams/mol4 H + O2 → 2 H2OWhich per mol is4 * 1 + 2 * 16 = 2 * 18 = 36 grams/molTwo moles of oxygen gas are consumed by one mole of methane and since and we started with 50 cm3 we consume 100 cm3 of oxygen at the same temperature and pressure by the methane alone. Whatever mass it turns out to be. (though at STP it would be 2*32/448=1/7 gram.EthyleneC2H4 by inspection we know2 C + 2 O2 → 2 CO2 two moles of oxygen4 H + O2 → 2 H2O one mole of oxygen.So 3 moles of Oxygen are consumed by each mole of ethylene. And since we started with 50 cm3 of ethylene we needed 150 cm3 of oxygen for the ethylene alone.Total100 + 150 = 250 cm3 Oxygen.At STP 5*32/448 = 5/14 gram.Though it's not needed for this problem it's always good to have an idea of what's going on.
Methane is more flammable than ethane because the methane molecules contain less of a number of atoms in each molecule than the methane has. Are there only the number of atoms in each molecule that determines how flammable a substance is?
Combustion is an interlocking chain of several free radical reactions. These reactions involve the breaking of bonds and the formation of other bonds to form the combustion products H2O and CO2, in your case from O2 and CH4 ( or C2H6).Notice that the number and types of bonds in methane are different from those in ethane. These bonds are broken at rates that depend on their bond strength. The global rate of the combustion is indeed related to all these bond breaking and bond forming reactions, but the rate equations become quite complex, even for a hydrocarbon as simple as methane.The flammability of an hydrocarbon can be measured experimentally, but these comparative flammability cannot be reliably calculated from the volatility of the hydrocarbon fuels, their C/H ratio, or the number of carbons in their linkages, or the hybridization of the constituent carbon atoms, or their branching, or their aromaticity. All these factors have been shown to affect the rate of combustion, yet the rate equations are too complex for ordinary engineering problems. However, these factors can be used to estimate comparative flammability, the temperature of the combustion, it’s speed of propagation, and the amount of soot it produces. Ordinary estimates based on any one or two of these factors ( such as the ones you mentioned ) could well be wildly off the mark.
How many moles of oxygen are needed for the complete combustion of 2 moles of methane?
For 1 mole of Methane:CH4 + 2 O2 —————-> CO2 + 2 H201 mole of Methane requires 2 mole of Oxygen.For 2 mole of Methane: Just make the number of mole of Methane 2 and balance the reaction2 CH4 + 4 O2 —————-> 2 CO2 + 4 H20Hence, 2 moles of Methane requires 4 moles of Oxygen for complete combustion.
What is the balanced chemical equation for the combustion of propane?
Propane ( C3H8 ) will burn completely when it combines with the oxygen (O2) in air to form carbon dioxide (CO2) and water (H2O). The equation looks like this: C3H8 + 5O2 ---> 3CO2 + 4H2O The heat generated in the exothermic reaction causes more and more propane to "break apart" and combine with oxygen in air to produce the end products carbon dioxide and water. This will continue until the concentration of propane in air falls below a "threshold" and not enough heat is generated to support the combustion of any remaining propane. There is enough oxygen in air in an open space to support the combustion of an extremely large volume of propane.
Why do longer hydrocarbon chains release more energy....?
The way fuel burns is in three stages. 1. First the Hydrogen is burned off the hydrocarbon releasing a little energy. It is oxidized to water vapor but the energy released is small. 2. Next the carbons become Carbon Monoxide with more energy released than the hydrogen burn. 3. Then the Carbon Monoxide is further oxidized to Carbon Dioxide with the most energy released. Its the breaking of the carbon-carbon bonds that releases the most oxygen. Of course that releases the most CO2 also.