Methane is the compound CH4, and burning it uses the reaction:
CH4 + O2 -> CO2 + H2O, which is rather exothermic. To find the heat released by burning a certain amount of the substance, you should look at the bond enthalpy of each compound, and then compare the values before and after the reaction. In methane, there are 4 C-H bonds, which have bond energy of 416 kj/mol, resulting in a total bond energy of 1664 kj/mol. O2 is 494 kj/mol. Therefore we have a total of 2080 kj/mol on the left side. On the right side we have CO2, which has 2 C=O bonds, each at 799 kj/mol each, resulting in 1598 kj/mol, and H2O has 2 O-H bonds, at 459kj/mol each, resulting in a total of 2516 kj/mol on the right hand side. Now, this may be confusing because the left hand side seems to have less heat than the right, but you just need to remember: making minus breaking, which results in a total change of 436kj/mol heat evolved.
Now it is a simple matter of find the mols of CH4 reacted, using n=m/mr.
n = 9.5/16.042 = 0.592195 mol
Therefore, if we reacted 0.592195 mol, and we produced 436 kj for one mol, the total amount of energy evolved was 436*<span>0.592195 kj, or 258.197 kj.</span>
Answer:
more rounded the grains are the more they have been moved around
Explanation:
Generally – the more rounded the grains are the more they have been moved around (i.e. the longer the length of time or distance they have moved). Angular grains cannot have travelled far
geolsoc.org.uk
Must be C because the liquid fuel is Codium nitrate to be used in rocket engines
The balanced chemical equation for the above reaction is as follows;
2Ca + O₂ --> 2CaO
stoichiometry of Ca to O₂ is 2:1
this means that 2 mol of Ca reacts with 1 mol of O₂.
If O₂ is the limiting reactant,
4 mol of O₂ should react with (4x2) - 8 mol of Ca
however only 7.43 mol of Ca is present. Therefore Ca is the limiting reactant.
7.43 mol of Ca reacts with - 7.43/2 = 3.715 mol of O₂
therefore there's excess O₂₂ remaining after the reaction
Since Ca is the limiting reactant, it is fully used up in the reaction and there is no Ca remaining after the reaction is completed.
