Câu 5:
\(m_{nặng.hơn}=m_{O\left(trong.oxit\right)}=0,16\left(g\right)\\Ta.có:m_{CO_2}+m_{H_2O}=m_{CO}+m_{H_2}+m_{O\left(nặng\right)}\\ \Leftrightarrow m_{CO_2}+m_{H_2O}=m_{CO}+m_{H_2}+0,16\\ m_{hhoxit}=m_{Cu}+m_{Fe}+m_{Al_2O_3}+0,16\\ \Leftrightarrow a=m_{Cu}+m_{Fe}+m_{Al_2O_3}=8,4-0,16=8,24\left(g\right)\\ n_{O\left(nặng\right)}=0,01\left(mol\right)=n_{CO}+n_{H_2}\\ \Rightarrow V=V_{H_2,CO\left(dktc\right)}=0,01.22,4=0,224\left(l\right)\\a= \)
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