Balance the equations for the formation of AlCl3(s) and CS2(l) from their elemen

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Calculate the reaction enthalpy for the formation of anhydrous aluminum chloride: 2Al(s) + 3Cl2(g) ---> 2AlCl3(s) from the following data: 2Al(s) + 6HCl(aq) ---> 2AlCl3(aq) + 3H2 (g) ?H° = -1049 kJ HCl(g) ---> HCl(aq) ?H° = -74.8 kJ H2(g) + Cl2(g) ---> 2HCl (g) ?H° = -185 kJ AlCl3(s) ---> AlCl3(aq) ?H° = -323 kJ Solution: 1) Let's examine each of the four equations in light of what needs to happen to it (in order to produce the target equation): eq. 1 ? this one remains unchanged. It gives us 2Al(s), which is what we want. The other substances will cancel out, as described below. eq. 2 ? this one will get multiplied by six in order to cancel the 6HCl(aq). eq. 3 ? this one gets multiplied by three. This gives us 3Cl2(g), which is what we want, and cancels out the six HCl(g) that was in eq. 2. It also cancels the 3H2(g) from eq. 1. eq. 4 ? this one gets flipped (to put AlCl3(s) on the right) and it gets multiplied by two. It also cancels the AlCl3(aq) from eq. 1.

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