1. Write and balance the equation.
2Mg + O2 ==> 2MgO
2. Convert what you have to mols.
a. mols Mg = grams/atomic mass = ?
b. mols O2 = grams/molar mass = ?
3. Using the coefficients in the balanced equation, convert mols Mg and mols O2 to mols of MgO
a. You likely will find that the mols MgO proeduced is NOT the same for both Mg and O2.
b. The SMALLER of the two is the correct value to use. The reagent (Mg or O2) producing the smaller number is the limiting reagent (LR).
Now for the mass of the EXCESS (ER),
1. Using the coefficients from the LR, convert that to mols of the ER. That will be mols ER USED.
2. Convert mols ER used to grams. grams = mols x molar mass = ?
3. Subtract grams ER used from grams ER initially from the problem. That will be grams ER remaining. That is the excess.
Post your work if you get stuck.
0.5 grms of the magnesium burnt in a closed vessel which contents 0.25grms oxygen.
1)which reaction limiting reagent?
2)find the mass of access reaction?
1 answer