Write the equation and balance it.
Convert 42 g Mg to moles. moles = grams/molar mass
Convert 45 g oxygen to moles the same way.
Using the coefficients in the balanced equation, convert moles Mg to mols MgO.
Same procedure, convert moles Oxygen to moles MgO.
The two answers you obtain probably will not be the same which means one of them is wrong. In limiting reagent problems, the correct answer is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
To find the grams of the product, use the smaller value from the previous step and g = moles x molar mass.
is that stoichiometry?
It is but it is a limiting reagent problem; therefore, you essentially do two stoichiometry problems to see which one is correct. The first one to see how much MgO would be produced if you had 42 g Mg and all the oxygen you needed and the second one to see how much MgO would be produced if you had 45 g oxygen and all of the Mg you needed.
Is it possible to do it without stoich?
Not really. However, it is possible to determine which is the limiting reagent, then use stoichiometry to calculate g MgO produced. The way you do the limiting reagent is as follows (BUT it really doesn't get around stoichiometry as you will see).
2Mg + O2 ==> 2MgO
moles Mg = 42/24.3 = about 1.73
moles O2 = 45/32 = about 1.4
Now we pick one reagent and convert to the other. Let's pick Mg since that is first. 1.73 moles Mg x (1 mol O2/2 moles Mg) = 1.83 moles Mg x 1/2 = about 0.9 moles O2. Do we have that much O2? Yes, we have 1.4; therefore, we know Mg is the limiting reagent. We then take that much Mg, convert to moles MgO, then to grams MgO.
What if we had picked oxygen first. So we convert oxygen to moles Mg as follows:
1.4 moles O2 x (2 moles Mg/1 mole O2) = 1.4 s 2 = 2.8 moles Mg needed. Do we have that much Mg? No, we have only 1.73; therefore, there isn't enough Mg to use all of the oxygen and we know Mg is the limiting reagent. I hope all of this helps.
this doesnt help...
swag money m8