1.
\(M_{CuO}\)= 64 + 16 = 80 (g)
\(M_{Fe_2O_3}\)= 56 .2+16.3 = 160 (g)
Theo đề bài cho:
\(m_{Fe_2O_3}=\frac{80}{100}\times50=40\left(g\right)\)
\(\rightarrow n_{Fe_2O_3}=\frac{40}{160}=0,25\left(mol\right)\)
\(m_{CuO}=\frac{20}{100}\times50=10\left(g\right)\)
\(\rightarrow n_{CuO}=\frac{10}{80}=0,125\left(mol\right)\)
Phương trình hóa học:
\(CuO+H_2\rightarrow Cu+H_2O\) (1)
0,125mol - 0,125 mol
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\) (2)
0,25mol- 0,75 mol
\(\Sigma n_{khí}H_2\)cần dùng: 0,125 + 0,75=0,875 mol
\(\Rightarrow V_{H_2}=0,875\times22,4=19,6l\)