Chemistry masterminds! I really need your help! If 17.50 g of KClO3 were used..

THERE ISNT ENOUGH TIME >.<

If 17.50 g of KClO3 were used in the following equation, what volume of
oxygen would be produced at 25 degree's C and 744 torr pressure?

2KClO3 ---------> 2KCl + 3O2

2moles of KClO3 gives 3 moles of O2
So, 17.5 g KClO3 will give [(3 x 32) / (2 x 122.5)] x 17.5 g of O2 = 6.85 g O2 ie 0.214 moles of O2
Now, Volume of O2 = 0.214 x 0.0821 x 298 / (744/760) = 5.35 liters