The molar mass off o2 is 32.0 g/mol. what mass in tye grams of 02 is required to react completely with 4.00 mol mg

The balanced reaction is: 

2Mg + O2 = 2MgO 

We are given the amount of  magnesium to be reacted with O2. This will be the starting point of the calculations.

4.00 g C3H8 ( 1 mol Mg/ 24.31 g Mg ) ( 1 mol O2 / 2 mol Mg ) ( 32.0 g O2 / 1 mol O2 ) = 2.63 g O2

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"