If 146.5 g of chromium (iii) chlorate decomposes what mass of oxygen can be produced

The Equation for this reaction is as follow,
                                2 Cr(ClO₃)₃    →    2 CrCl₃  +  9 O₂

According to eq,

       604.68 g (2mole) of Cr(ClO₃)₃ produces  =  288 g (9 moles) of O₂
So,
                  146.5 g of Cr(ClO₃)₃ will produce  =  X g of O₂

Solving for X,
                              X  = (146.5 g × 288 g) ÷ 604.68 g

                              X  =  69.77 g of O₂

It is called a neutralization reaction

The correct answer is c