Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a, PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
___0,1__0,05__0,1 (mol)
b, Ta có: \(V_{O_2}=0,05.22,4=1,12\left(l\right)\)
c, Ta có: \(m_{H_2o}=0,1.18=1,8\left(g\right)\)
Bạn tham khảo nhé!
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(2H_2+O_2\underrightarrow{t^0}2H_2O\)
\(0.1......0.05....0.1\)
\(V_{O_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(m_{H_2O}=0.1\cdot18=1.8\left(g\right)\)
a/ PTHH: \(2H_2+O_2\overset{t^o} 2H_2O\)
b/ \(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\)
Theo pt: \(2:1:2\)
\(\to n_{O_2}=0,05(mol)\)
\(\to V_{O_2}=0,05.22,4=1,12(l)\)
b/ \(n_{H_2O}=0,1(mol)\)
\(\to m_{H_2O}=0,1.18=1,8(g)\)
