Question

Câu 10. Nhiệt phân m gam KClO 3 sau pư thu được 2,688 lít khí O2 (đktc) tìm m, biết hao hụt 20% 

Answer 1

2KClO3-to>2KCl+3O2

0,08--------------------0,12 mol

n O2=2,688\22,4=0,12 mol

H=20%

=>m KClO3tt=0,08.122,5.100\20=49g

Answer 2

\(n_{O_2}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\\ 2KClO_3\underrightarrow{^{to}}2KCl+3O_2\\ n_{KClO_3\left(LT\right)}=\dfrac{2}{3}.0,12=0,08\left(mol\right)\\ Hao.hụt.80\%.Nên:n_{KClO_3\left(TT\right)}=0,08:\left(100\%-20\%\right)=0,1\left(mol\right)\\\Rightarrow m=m_{KClO_3\left(TT\right)}=122,5.0,1=12,25\left(g\right)\)