\(n_{Fe2O3}=\frac{24}{160}=0,15\left(mol\right)\)
\(n_{H2O}=\frac{1,8.10^{23}}{6,02.10^{23}}=0,3\left(mol\right)\)
a. \(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
0,1_____ 0,3_____0,2_____0,3
\(n_{H2}=0,3\left(mol\right)\)
\(\rightarrow V_{H2}=0,3.22,4=6,72\left(l\right)\)
b. \(n_{Fe2O3_{Du}}=0,15-0,1=0,05\left(mol\right)\)
Rắn gồm 0,2mol Fe và 0,05mol Fe2O3
\(\rightarrow m=0,2.56+0,05.160=19,2\)
c. \(\%m_{Fe2O3_{pu}}=\frac{0,1}{0,15}=66,67\%\)
\(n_{Mg}=\frac{9,6}{24}=0,4\left(mol\right)\)
a. \(2Mg+O_2\rightarrow2MgO\)
MgO: Magie oxit
\(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)
\(\rightarrow m_{MgO}=0,4.40=16\left(g\right)\)
b. \(n_{O2}=\frac{nMg}{2}=0,2\left(mol\right)\)
\(\rightarrow V_{O2}=0,2.22,4=4,48\left(l\right)\)
\(\rightarrow V_{KK}=\frac{4,48}{20\%}=22,4\left(l\right)\)