Ta có
\n\n\\(n_{Zn}=\\frac{9,75}{65}=0,15\\left(mol\\right)\\)
\n\n\\(PTHH:2Zn+O_2\\rightarrow2ZnO\\)
\n\n________0,015___0,075____
\n\n\\(\\Rightarrow\\Sigma n_{O2}=0,075.5=0,375\\left(mol\\right)\\)
\n\nDo bị hao hụt 40% nên:
\n\n\\(n_{O2}=0,035-0,035.40\\%=0,225\\left(mol\\right)\\)
\n\n\\(2KClO_3\\rightarrow2KCl+3O_2\\)
\n\n\\(\\Rightarrow m_{KClO3}=0,15.12,5=18,375\\left(g\\right)\\)
\n