Question

Ai giải giúp mình với

Answer 1

Ta có: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{40\cdot60\%}{160}=0,15\left(mol\right)\\n_{CuO}=\dfrac{40\cdot40\%}{80}=0,2\left(mol\right)\end{matrix}\right.\)

PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)

                  0,15____0,45___0,3____0,45  (mol)

            \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

                 0,2___0,2___0,2___0,2  (mol)

Ta có: \(\left\{{}\begin{matrix}m_{Fe}=0,3\cdot56=16,8\left(g\right)\\m_{Cu}=0,2\cdot64=12,8\left(g\right)\\V_{H_2}=\left(0,2+0,45\right)\cdot22,4=14,56\left(l\right)\end{matrix}\right.\)