a) nZn=6,5÷65=0,1(mol)
ADCT: CM=n/V → V=n/CM
→V=0,1÷2=0,05(l)
b) theo ý a) ta có: nZn=0,1(mol)
PTHH: Zn(0,1) + 2HCl → ZnCl2 + H2(0,1)
Theo pt ta có: nH2=0,1(mol)
→ VH2=0,1×22,4=2,24(l)
PTHH: Zn + 2HCl -> ZnCl2 + H2
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo PTHH và đề bài, ta có: \(\left[{}\begin{matrix}n_{HCl}=2.n_{Zn}=2.0,1=0,2\left(mol\right)\\n_{H_2}=n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)
a) Thể tích dd HCl:
\(V_{ddHCl}=\dfrac{n_{HCl}}{C_{MddHCl}}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)
b) Thể tích khí H2 thu được (đktc):
\(V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
Ta co pthh
Zn + 2HCl \(\rightarrow\)ZnCl2 + H2
Theo de bai ta co
nZn=\(\dfrac{6,5}{65}=0,1mol\)
a,Theo pthh
nHCl=2nZn=2.0,1=0,2 mol
Ta co nong do cua dd HCl la
CM=\(\dfrac{nHCl}{VHCl}\Rightarrow VHCl=\dfrac{nHCl}{CM}=\dfrac{0,2}{2}=0,1l\) = 100 l
b, Theo pthh
nH2=nZn=0,1 mol
\(\Rightarrow VH2=0,1.22,4=2,24l\)
