# 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?

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Consider the following reaction: #2Mg(s) + O_2(g) -> 2 MgO(s)# #DeltaH = -1204 kJ# .

Consider the following reaction:

##### 1 Answer

#### Answer:

a) -87.9 kJ

b) 15.7 g

#### Explanation:

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

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

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 *Mg and Enthalpy are related in a 2:1 ratio.* Keeping this in mind:

So, the enthalpy change is **released** is

b) In the next step we do the same thing, just backwards. First, let's talk about the relationship between *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:

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

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

And you're done.

Hope that helped :)