\(m_{O_2}=3.33-2.13=1.2\left(g\right)\)
\(n_{O_2}=\dfrac{1.2}{32}=0.0375\left(mol\right)\)
\(V_{O_2}=0.0375\cdot22.4=0.84\left(l\right)\)
Bảo toàn O :
\(n_{H_2O}=2n_{O_2}=2\cdot0.0375=0.075\left(mol\right)\)
Bảo toàn H :
\(n_{HCl}=2n_{H_2O}=0.075\cdot2=0.15\left(mol\right)\)
\(V_{dd_{HCl}}=\dfrac{0.15}{2}=0.075\left(l\right)\)
a, Bảo toàn khối lượng ta có: $m_{O_2}=1,2(g)\Rightarrow n_{O_2}=0,0375(mol)\Rightarrow V_{O_2}=0,84(l)$
b, Ta có: $n_{O}=2.n_{O_2}=0,075(mol)$
$\Rightarrow n_{HCl}=2.n_{O}=0,15(mol)$
$\Rightarrow V_{HCl}=0,075(l)$
a, nO2- = 3,33 - 2,13 / 16 = 1,2 => nO2 = 1,2 /2 = 0,6 (mol ) => V = 0,6x22.4 = 13,44 (l)
b, ne = 1,2x2 = 2,4 => nCl- = 2,4 => V = n/Cm = 1,2 (l)
