Question

Calculate the mass of O2 produced after 20.0 grams of KClO3 have reacted?

Answer 1

Answer:

$RFM \: of \: potassium \: chlorate = (39 + 35.5 + (16 \times 3)) = 122.5 \: g \\ 122.5 \: g \: of \: potassium \: chlorate \: produce \: (16 \times 3) \: g \: of \: oxygen \\ 20.0 \: g \: of \: potassium \: chlorate \: will \: produce \: (\frac{(20.0 \times 16 \times 3)}{122.5} ) \: g \\ = 7.84 \: g \: of \: oxygen$