How would you calculate the amount of heat transferred when 3.55 g of Mg(s) react at constant pressure? How many grams of MgO are produced during an enthalpy change of -234 kJ?

This is simply a stoichiometry problem, but involving the enthalpy as well. The key to these problems is that you simply treat the enthalpy like another product. So, in this case, we'd interpret the reaction as:

For every #2# mol of #Mg# that react with 1 mol of #O_2#, you get #2# mol of #MgO# and release (which you know because of the negative sign of the enthalpy) #1204 kJ# of heat.

a) Now, all we need to do is deal with this like we would a regular stoichiometry problem. So we're given a mass of #Mg(s)#. Let's go ahead and convert that to moles, so we can work with it in the reaction:

#(3.55 g)/(24.31 g/(mol)) =# 0.146 mol Mg

Now, we need to find enthalpy. Remember, we just treat it like another product. We know that for every 2 mol Mg, we get an enthalpy release of #1204 kJ#. So, therefore, Mg and Enthalpy are related in a 2:1 ratio. Keeping this in mind:

#DeltaH = 0.146/2*-1204 kJ = -87.91 kJ#

So, the enthalpy change is #-87.91 kJ#, or the amount of heat released is #87.91 kJ#.

b) In the next step we do the same thing, just backwards. First, let's talk about the relationship between #MgO# and enthalpy. If you go back to the reaction, you'll see that the relationship is that for every 2 mol of MgO, you release (again, which we know because of the negative sign) 1204 kJ of heat

However, we didn't release 1204 kJ of heat; we released 234 kJ. So, we're going to need to get a ratio, which we can then play with in the reaction:

#(-234)/-1204 = 0.194#.

Now all we need to do is simply multiply 0.194 times 2 to find moles of #MgO#:

#0.194 * 2 =# 0.389 mol Mg

Keep in mind, however that this is moles of Mg. You need grams. So, we will have to multiply by the molar mass of #MgO#:

#(0.389 mol)(40.31 g/(mol)) = #15.7 g MgO.

And you're done.

Hope that helped :)