Question

Nung 12,25 gam KClO 3 sau pư thu được 2,016 lít O 2 (đktc). Tính hiệu suất pư?

Answer 1

2KClO3-to>2KCl+3O2

0,06-----------------0,09  mol

n O2=2,016\22,4=0,09 mol

=>H =0,06.122,5\12,25 .100=60%

Answer 2

\(n_{O_2\left(TT\right)}=\dfrac{2,016}{22,4}=0,09\left(mol\right)\\ n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\\ 2KClO_3\underrightarrow{^{to}}2KCl+3O_2\\ n_{O_2\left(LT\right)}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ \Rightarrow H=\dfrac{0,09}{0,15}.100=60\%\)