How many grams of o2 is produced if you start with 12.5 grams of naclo3? 2naclo3(s) --> 2nacl(s) + 3o2(g) a) 5.64 grams o2 b) 23.1 grams o2 c) 0.35 grams o2 d) 36.8 grams o2

According to balanced chemical equation 212 g (2 moles) of NaClO3 on complete consumption produces 96 g (3 moles) of O2. So, using this as a reference point the amount of O2 produced by consuming 12.5 g of NaClO3 will be calculated as,

212 g of NaClO3 produced / 12.5 g NaClO3 will produce = 96 g of O2 / x

Solving for x,

x = 12.5 g /212 g x 96 g

x = 5. 66 g

So, the correct answer is A.

Alpha particles have positive charge, beta particles have negative charge, and answer 4 just doesn’t make sense. thus, you are left with only one possible answer.