2 H2 + O2 — > 2 H2O
Methane + Oxygen = carbon dioxide + water vapour
That would refer to a mixture of ethanol alcohol and methanol alcohol. Also referred to as “denatured” alcohol. This is typically alcohol that is supposed to be used for cleaning or solvent purposes; methanol is added to make it poisonous to drink so that the manufacturer and hardware stores don't need to pay the alcohol tax.
What is the chemcial equation for the burning of a candle?
A complete and accurate equation cannot be given. Candle wax consists of a number of heavy hydrocarbon compounds ranging from C15 to C30 depending on the type of candle. A party candle will have fewer, smaller compounds (low density), while a large Church candle will have a much higher density due to the larger molecules of wax. When you see a candle burning, the amount of oxygen that can react with the wax vapour from the wick, is insufficient. The reaction therefore will not give complete combustion and results in a smoky, yellow flame indicating that Carbon 'C' (as Soot) is formed. Also, the shortage of O2 will give less CO2, some CO (monoxide) and water vapour (H2O), together with heat and light.
What is the chemcial equation for the burning of a candle?
A complete and accurate equation cannot be given. Candle wax consists of a number of heavy hydrocarbon compounds ranging from C15 to C30 depending on the type of candle. A party candle will have fewer, smaller compounds (low density), while a large Church candle will have a much higher density due to the larger molecules of wax. When you see a candle burning, the amount of oxygen that can react with the wax vapour from the wick, is insufficient. The reaction therefore will not give complete combustion and results in a smoky, yellow flame indicating that Carbon 'C' (as Soot) is formed. Also, the shortage of O2 will give less CO2, some CO (monoxide) and water vapour (H2O), together with heat and light.
CH2=CH-CH2-CH3 is But-1-ene.It's Molecular Formula C4H8. This on complete combustion gives.C4H8+6O2———->4CO2+4H2O.If the Molecular Formula of the Hydro carbon is CxHy then x- moles of Carbon dioxide and y/2 moles of Water will be formed.
Jayank's answer is perfectly good for introductory chemistry exams, but the reality of the situation is more complicated. The combustion of methanol (or any other hydrocarbon) is not a single step reaction, involving a single equation. There are multiple steps and often multiple timescales in these kinds of reactions.Combustion starts with a system of energetic fuel molecules, created by heating the system up such that the molecules vibrate intensely. At some point, the molecule vibrates so violently, that one of the hydrogen atoms gets knocked off,CH3OH + energy --> C*H2OH + H*producing the hydroxymethyl radical, C*H2OH, and a hydrogen atom, where I am using an asterisk to indicate the position of the unpaired electron. These hydrogen atoms react with the molecular oxygen to eventually form water,4 H* + O2 --> 2 H2OWhen the C*H2OH radical collides with an oxygen molecule, one of the oxygen atoms attaches to where that first hydrogen came off,C*H2OH + O2 --> CH2(OO*)OHforming a hydroxymethyl peroxyl radical, in which the unpaired electron moves to one of the oxygen atoms. The oxygen atom with the unpaired electron then plucks off a hydrogen from somewhere else in the molecule. In this case, removing the hydrogen from the hydroxy group (OH) is energetically favorable, and the dominant pathway isCH2(OO*)OH --> CH2(OOH)O*followed by an elimination of the hydroperoxyl radical (HO2),CH2(OOH)O* -->CH2O + HO2The hydroperoxyl radical is relatively stable and long-lived, but eventually reacts with hydrogen atoms to form water. We are then left with the formaldehyde molecule, CH2O. At this point, the process essentially repeats itself:CH2O + energy --> C*HO + H4H + O2 -- > 2H2OC*HO + O2 --> C(OO*)HO --> C*(OOH)OThis time, there are two options: direction elimination of HO2C*(OOH)O --> HO2 + COor elimination of the highly reactive hydroxyl (OH) radicalC*(OOH)O --> CO2 + OHYou can see that basically either of these reactions will yield carbon dioxide (CO2) and water (H2O) in subsequent steps. Thus, to summarize the whole process, we can write2CH3OH + 3O2 --> 2CO2 + 4H2O
First, we need to know the formulas for the chemicals involved.Magnesium, I guess, means elemental magnesium - which is typically just a metal. This means that it’s possible for single atoms to react, so the formula is Mg_1, or just Mg.Oxygen is commonly a diatomic gas, meaning that it contains 2 atoms per molecule - hence a formula of O_2.Magnesium oxide consists of two ions - magnesium ions (which form when magnesium loses 2 electrons), and oxide ions (which form when an oxygen atom gains 2 electrons). This 2:2 ratio of electrons exchanged means a 2:2 ratio of atoms needed to form a neutral ionic compound, which can be simplified to 1:1 - hence, Mg_1O_1 or MgOHence, chemical equation isMg + O_2 -> MgO.Now, we need to balance the equation.Left: 1 Mg, 2 O.Right: 1 Mg, 1 O.So let’s double the MgO:Mg + O_2 -> 2MgOLeft: 1 Mg, 2 ORight: 2 Mg, 2 O.So let’s double the elemental Mg:2Mg + O_2 -> 2MgOLeft: 2 Mg, 2 ORight: 2 Mg, 2 OAnd our equation is balanced:2Mg + O_2 = 2MgO
All hydrocarbons completely combust with dioxygen to give carbon dioxide and water…And thus propane and dioxygen gives carbon dioxide and water. And we can write the balanced symbol equation, following the usual rigmarole of (i) balance the carbons as carbon dioxide, (ii) then balance the hydrogens as water, and (iii) THEN balance the oxygens…[math]C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(l) + \Delta_{1}[/math]Even-numbered alkanes require a half-integral dioxygen coefficient…i.e. for ethane..[math]C_{2}H_{6}(g) + \frac{7}{2}O_{2}(g) \rightarrow 2CO_{2}(g) + 3H_{2}O(l) + \Delta_{2}[/math]How would [math]\Delta_{1}[/math] compare to [math]\Delta_{2}[/math]?