\(2H_2+O_2\rightarrow2H_2O\)
a: \(n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(n_{O_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
Vì \(\dfrac{n_{H_2}}{2}< \dfrac{n_{O_2}}{1}\)
nên O2 dư
b: \(n_{H_2O}=0.05\left(mol\right)\)
\(m_{H_2O}=0.05\cdot18=0.9\left(g\right)\)
a.\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH : 2H2 + O2 -> 2H2O
0,05 0,025 0,05
Xét tỉ lệ \(\dfrac{0,05}{2}< \dfrac{0,1}{1}\) => H2 đủ , O2 dư
\(m_{O_2\left(dư\right)}=\left(0,1-0,025\right).32=2,4\left(g\right)\)
b. \(m_{H_2O}=0,05.18=0,9\left(g\right)\)
c) PTHH : 3Fe + 2O2 -> Fe3O4
0,025 0,0125
\(n_{Fe_3O_4}=0,0125.232=2,9\left(g\right)\)
