calculate the enthalpy of formationof solid MgOH2, given the following data:

Question

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?

Answer 1

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.

Answer 2

the units are all kj/mol. and no i did not, the second equation is supposed to be endothermic.

Answer 3

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.

Answer 4

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?