a) \(n_{H_2}=\dfrac{28}{22,4}=1,25\left(mol\right)\)
PTHH: \(2H_2O\underrightarrow{đp}2H_2+O_2\)
1,25<---1,25
=> m = 1,25.18 = 22,5 (g)
b)
PTHH: \(H_2+Cl_2\underrightarrow{as}2HCl\)
1,25---------->2,5
=> \(n_{HCl\left(thu.được\right)}=\dfrac{2,5.90}{100}=2,25\left(mol\right)\)
=> mHCl = 2,25.36,5 = 82,125 (g)
a) $2H_2O \xrightarrow{điện\ phân} 2H_2 + O_2$
$n_{H_2} = \dfrac{28}{22,4} = 1,25(mol)$
$n_{H_2O} = n_{H_2} = 1,25(mol) \Rightarrow m = 1,25.18 = 22,5(gam)$
b) $H_2 + Cl_2 \xrightarrow{ánh\ sáng} 2HCl$
$n_{HCl} = 2n_{H_2} = 2,5(mol)$
$n_{HCl\ thu\ được} = 2,5.90\% = 2,25(mol)$
$m_{HCl} = 2,25.36,5 = 82,125(gam)$
