Theo gt ta có: $n_{H_2}=0,15(mol)$
a, $2H_2+O_2\rightarrow 2H_2O$
b, Ta có: $n_{O_2}=\frac{1}{2}.n_{H_2}=0,075(mol)\Rightarrow V_{O_2}=1,68(l)$
c, Ta có: $n_{H_2O}=0,15(mol)\Rightarrow m_{H_2O}=2,7(g)$
PTHH: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{H_2O}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{H_2O}=0,15\cdot18=2,7\left(g\right)\end{matrix}\right.\)
Theo gt ta có: nH2=0,15(mol)
a, 2H2+O2 to--> 2H2O
nO2 =0.5.nH2=0,075(mol)⇒VO2=1,68(l)
c, Ta có: nH2O=0,15(mol)⇒mH2O=2,7(g)
a. 2H2 + O2 ---to---> 2H2O.
b. Ta có: nH2=3,36/22,4=0,15(mol)
Theo PT, ta có: nO2=1/2 . 0,15=0,075(mol)
=> VO2=0,075.22,4=1,68(l)
c. Theo PT, ta có: nH2O=nH2=0,15(mol)
=> mH2O=0,15.18=2,7(g)
