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a) Zn + 2HCl →ZnCl2 + H2
b) nZn = 6,5/65 = 0,1 mol . Theo tỉ lệ pư => nH2 = nZn = nZnCl2 =0,1 mol <=> VH2(đktc) = 0,1.22,4 = 2,24 lít.
c) mZnCl2 = 0,1 . 136 = 13,6 gam
d) nHCl =2nZn = 0,2 mol => mHCl = 0,2.36,5= 7,3 gam
Cách 2: áp dụng định luật BTKL => mHCl = mZnCl2 + mH2 - mZn
<=> mHCl = 13,6 + 0,1.2 - 6,5 = 7,3 gam
Theo gt ta có: $n_{Zn}=0,1(mol)$
a, $Zn+2HCl\rightarrow ZnCl_2+H_2$
b, Ta có: $n_{H_2}=0,1(mol)\Rightarrow V_{H_2}=2.24(l)$
c, Ta có: $n_{HCl}=2.n_{Zn}=0,2(mol)\Rightarrow m_{HCl}=7,3(g)$
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=1\left(mol\right)\\n_{H_2}=0,5\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{1\cdot36,5}{20\%}=182,5\left(g\right)\\V_{H_2}=0,5\cdot22,4=11,2\left(l\right)\end{matrix}\right.\)
a+b+c) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
d) PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PTHH: \(n_{H_2}=n_{Cu}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)
\(a,n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1--->0,2------>0,1----->0,1
\(\rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{10\%}=73\left(g\right)\\ b,m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\ V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(nZn=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(Zn+2HCl->ZnCl_2+H_2\)
0,1 0,2 0,1 0,1 (mol)
\(mHCl=0,2.36,5=7,3\left(g\right)\)
=> \(mddHCl=\dfrac{7,3.100}{10}=73\left(g\right)\)
mZnCl2 = 0,1 . 136 = 13,6 )g_
VH2 = 0,1 . 22,4 = 2,24 (l)
a, \(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{Zn}=0,4\left(mol\right)\Rightarrow m_{H_2}=0,4.2=0,8\left(g\right)\)
b, \(2H_2+O_2\underrightarrow{^{t^o}}2H_2O\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=0,2\left(mol\right)\Rightarrow V_{O_2}=0,2.22,4=4,48\left(l\right)\)
\(\Rightarrow V_{kk}=5V_{O_2}=22,4\left(l\right)\)
a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(n_{HCl}=0,2.1=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) => Zn dư, HCl hết
PTHH: Zn + 2HCl --> ZnCl2 + H2
__________0,2-------------->0,1
=> VH2 = 0,1.22,4 = 2,24(l)
b)
PTHH: 2H2 + O2 --to--> 2H2O
______0,1->0,05
=> mO2 = 0,05.22,4 = 1,12 (l)
$a\big)$
$n_{Zn}=\dfrac{3,25}{65}=0,05(mol)$
$Zn+2HCl\to ZnCl_2+H_2$
Theo PT: $n_{ZnCl_2}=n_{Zn}=0,05(mol)$
$\to m_{ZnCl_2}=0,05.136=6,8(g)$
$b\big)$
Theo PT: $n_{HCl}=2n_{Zn}=0,1(mol)$
$\to V_{dd\,HCl}=\dfrac{0,1}{0,5}=0,2(l)=200(ml)$
`Zn + 2HCl -> ZnCl_2 + H_2`
`0,05` `0,1` `0,05` `(mol)`
`n_[Zn]=[3,25]/65=0,05 (mol)`
`a)m_[ZnCl_2]=0,05.136=6,8(g)`
`b)V_[dd HCl]=[0,1]/[0,5]=0,2(M)`