calculate the enthalpy of formationof solid MgOH2, given the following data:
2Mg(s) + O2(g)-> 2MgO(s) heat:-1203.6
Mg OH2(s) -> MgO(s) +H2O(l) heat: 37.1
2H2(g)+ O2(g) -> 2H2O(l) heat: -571.7
what equation would they want the three equations to develop into and which ones would you flip/multiply?
First, magnesium hydroxide is Mg(OH)2
Second, if #1 give off heat then delta H = -1203.6 what. Joules. kJ. And is that kJ/mol or kJ for the reaction as written.
Third, since you show heat given off for all three equations, did you omit the - sign for #2 reaction; i.e., should that be -37.1 and what are the units.
the units are all kj/mol. and no i did not, the second equation is supposed to be endothermic.
ok. got it. If #2 is endothermic, then you should have written it as
Mg(OH)2(s) + heat ==> MgO(s) + H2O(l)
and I am going with Mg(OH)2 and not MgOH2
Use equation #1 as is but if delta H is kJ/mol, then you must multiply by 2 to obtain delta H for the reaction.
Reverse #2 and double it. Also write the reaction as doubled.
Use equation 3 in the same direction but double it if kJ/mol for there are 2 mols. Check my thinking. I wrote this out on paper and I have
2 Mg(s) + 2O2(g) + 2H2(g) ==> 2Mg(OH)2(s)
delta H for the final is the sum of the delta Hs for each as rewritten. Check my thinking. You must make certain that the delta Hs you have written are kJ/mol for that makes a big difference whether #1 and #3 are doubled or not.
Nitromethane (CH3NO2) burns in air to produce significant amounts of heat.
2CH3NO2(l)+3/2O2(g)→2CO2(g)+3H2O(l)+N2(g)
ΔHorxn = -1418 kJ
Part A
How much heat is produced by the complete reaction of 5.82 kg of nitromethane?
To calculate the enthalpy of formation of solid MgOH2, we need to combine the given equations such that they cancel out the common reactants and products. This will leave us with the desired equation of interest, which can then be used to calculate the enthalpy of formation.
The desired equation is:
2Mg(s) + 2H2(g) + O2(g) -> 2MgOH2(s)
To obtain this equation, we need to manipulate the given equations appropriately. Let's analyze each equation and see what modifications are required:
1. 2Mg(s) + O2(g) -> 2MgO(s) ΔH = -1203.6 kJ
This equation contains MgO, which is one of the desired products. However, we need to flip the equation, so it becomes:
2MgO(s) -> 2Mg(s) + O2(g) ΔH = +1203.6 kJ
2. Mg OH2(s) -> MgO(s) + H2O(l) ΔH = +37.1 kJ
This equation contains MgO as well, which is also one of the desired products. However, we need to multiply this equation by 2 so that the number of MgO matches the desired equation:
2MgOH2(s) -> 2MgO(s) + 2H2O(l) ΔH = +74.2 kJ
3. 2H2(g) + O2(g) -> 2H2O(l) ΔH = -571.7 kJ
This equation contains H2O, which is one of the desired products. However, we need to flip this equation so that it turns into:
2H2O(l) -> 2H2(g) + O2(g) ΔH = +571.7 kJ
Now, let's add the three modified equations together to cancel out the common reactants and products, leaving us with the desired equation:
2MgO(s) + 2H2O(l) -> 2MgOH2(s) ΔH = -1203.6 kJ + 74.2 kJ + 571.7 kJ
Simplifying the equation yields:
2MgO + 2H2O -> 2MgOH2 ΔH = -557.7 kJ
Thus, the three given equations combine to form the desired equation 2MgO + 2H2O -> 2MgOH2, with ΔH = -557.7 kJ.