Consider these reactions, where m represents a generic metal. 2m(s)+6hcl(aq)â¶2mcl3(aq)+3h2(g)î´h1=â768.0 kj hcl(g)â¶hcl(aq) î´h2=â74.8 kj h2(g)+cl2(g)â¶2hcl(g) î´h3=â1845.0 kj mcl3(s)â¶mcl3(aq) î´h4=â198.0 kj use the given information to determine the enthalpy of the reaction 2m(s)+3cl2(g)â¶2mcl3(s)

I really had a hard time understanding the data written in this problem. Luckily, I found a similar problem from another website shown in the attached picture. In order to solve this problem, you have to manipulate the reactions in the numbered list so that they add up to the reaction: 2M(s) + 3Cl₂ --> 2 MCl₃ (s). To do this, add up the reactions together such that both substances that appear in the left and right side of the reaction would cancel out. To be consistent, if you want to reverse the reaction, change the sign of H to the opposite sign. Also, if you multiply the whole reaction with 3 to completely cancel out like compounds, multiply the value of H with the same value. The addition of the reactions is as follows:

2M(s) + 6HCl(aq) --> 2 MCl₃(aq) + 3H₂(g)                  H = -748 kJ
2MCl₃(aq)            --> 2 MCl₃(s)                                   H = +317(2) kJ
6HCl(g)                --> 6HCl(aq)                                    H = -74.8(6) kJ
3H₂(g) + 3Cl₂(g)  --> 6HCl(g)                                      H = -1845(3) kJ
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2M(s) + 3Cl₂       --> 2 MCl₃ (s)                                  H = -4,601.8 kJ

hola

Explanation:Añada cinco gotas al 2 por ciento de tintura de yodo de farmacopea EE. UU. (U.S.P., por sus siglas en inglés) por cada litro de agua limpia. Para el agua turbia añada diez gotas y deje la solución reposar durante 30 minutos por lo menos.