How many grams of o2 are necessary to react with 212 g of mg 2mg+o2➡2mgo?

We are given an equation 2Mg+O2-> 2MgO and a starting chemical Mg of about 212 g. In order to solve for the amount of O2 needed, we need the molecular weight of Mg and O2.
Molecular weight:
Mg=24.305 g/mol
O2=16(2)=32 g/mol

Note that for every 1 mol of O2, the amount of Mg must be 2 mol.
So,
 g O2 = 212 g Mg x1mo Mgl/24.305 g Mg x1mol O2 /2 mol Mg x 32 g O2/mol O2

gO2=139.56 g

Therefore, 139.56 g of O2 is needed for every 212 g Mg.

m=9

step-by-step explanation:

7m+8=71 subtract   8 from 71 which is 63 than divide it by 7m   equal 9.