2M(s) + 6HCl(aq)—–> 2MCl3(aq) + 3H2(g) H1 = -885.0 kJ
HCl(g)—–>HCl(aq) H2 =-74.8 kj
H2(g) + Cl2(g)——> HCl(g) H3 = -1845.0 kj
MCl3(aq) ——> MCl3(s) H4 = -161.0 Kj
Use the information above to determine the enthalpy of the following reaction.
2M(s) + 3Cl2(g) —–> 2MCl3(s) H = __________ kJ
Ok what i was doing was Add 1*Reaction 1 + 2*Reaction 4 – 6*reaction 2 – 3*reaction 3 but it still says i was wrong, and i dont understand what im doing worng
friend u gone wrong in subtracting the reaction there is no need of subtraction here
just add the 1st reaction + 6 times the second reaction + 3 times the third reaction +2 times the fourth one
giving the answer to be -7190.85kJ
Since 2M(s) is in the reactants of the goal equation, we know we’ll use equation #1 in the forward direction.
Since 2MCl3(s) is the products of the goal equation, we know we’ll use equation #4 in the reverse direction, and multiplied by a factor of 2.
To get 6HCl(g) and cancel out 6HCl(aq), we need to use reaction #2 multiplied by a factor of 6.
Equation #3 is useful for cancelling out HCl(g) and H2(g), but we need to multiply it by a factor of 3.
Than add up equation 1 thru three in positive number and subtract the four equation for you answer.