Limiting reagent problems are done alike.
1. Write and balance the equation.
2Mg + O2 ==> 2MgO
2. Convert grams to moles.
a. 42 grams oxygen/32 = 1.31 moles O2.
b. 42 grams Mg/24.305 = 1.73 moles Mg.
3. Using the coefficients in the balanced equation, convert moles in 2a and 2b to moles of the product.
a. from O2: moles MgO = 1.31 moles O2 x (2 moles MgO/1 mole O2) = 1.31 x 2 = 2.62 moles MgO formed.
b. from Mg: moles MgO = 1.73 moles Mg x (2 moles MgO/2 moles Mg) = 1.73 x 2/2 = 1.73 moles MgO formed.
c. USUALLY, moles of the product from a and from b are different, as they are in this case. You know both answers can't be right. Which one is right? The smaller value is ALWAYS the correct value AND THE MATERIAL PRODUCING THAT NUMBER IS THE LIMITING REAGENT (in this case Mg).
4.a. If the problem asks (this one does in the last part) for grams MgO formed, that's just the smaller value x molar mass; i.e., 1.73 moles MgO x molar mass MgO = grams MgO.
b. If the problem asks which reactant is left over, you have that information in step 3. The limiting reagent is Mg; therefore, oxygen must be the material left.
c. Some of these problems (this one doesn't) asks for how much of the "other" material is left over? You do that exactly as in step 3.
moles O2 used = 1.73 moles Mg x (1 mole O2/2 mols Mg) = 1.73/2 = 0.865 mols O2 used. We had 2.62 initially - 0.865 moles used = 1.75 moles O2 remaining un-reacted. You can conert that to grams by 1.75 g O2 x molar mass O2 = ??
You need to go through this again and assign good numbers to each part. I've rounded here and there and estimated at some other spots. Put this procedure somewhere so you can refer to it later.
When one pound of gasoline (made up of compounds of hydrogen and carbon) is burned in an automobile approximately 3 pounds of carbon dioxide, CO2, is given off. Carbon dioxide is one of the gases contributing to global warming. What information from this experiment helps explain how one pound of gasoline can give off approximately 3 times an much CO2?