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Consider these reactions, where M represents a generic metal.?

2M(s) + 6HCl(aq)—–> 2MCl3(aq) + 3H­2(g) H1 = -885.0 kJHCl(g)—–>HCl(aq) H2 =-74.8 kjH2(g) + Cl2(g)——> HCl(g) H3 = -1845.0 kjMCl3(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 = __________ kJOk 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

2 Answers

A
Alycia Zemlak

Feb 20, 2025

friend u gone wrong in subtracting the reaction there is no need of subtraction herejust add the 1st reaction + 6 times the second reaction + 3 times the third reaction +2 times the fourth onegiving the answer to be -7190.85kJhess law... Show More
A
Alycia Zemlak

Feb 20, 2025

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.... Show More

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