Question

If 12.01 g of Mg metal and 6.56 g of O2 gas are reacted,
1) How many moles of MgO can be produced from the given mass of Mg?
2) How many moles of MgO can be produced from the given mass of O2?
3) From the above calculations, determine which of the reactants is the limiting reagent.
Recall the balanced reaction is:
2 Mg (s) + O2 (g) 2 MgO (s)
Enter the number of moles of MgO from 12.01 of Mg here:

Enter the number of moles of MgO from 6.56 of O2 here:

The limiting reagent is?: Mg O2

Answer 1

2Mg + O2 ==> 2MgO

mols Mg = grams/atomic mass = 12.01/24.3 = ?
mols O2 = grams/molar mass = 6.56/32 = ?

Using the coefficients in the balanced equation, convert mols Mg to mols MgO produced. That's #1.
#2. Convert mols O2 to mols MgO.
#3. In limiting reagent problems the SMALLER value will be the correct number of mols. The reagent responsible for the value is called the limiting reagent. It will be used up completely and the other one is the excess reagent.