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a) \(n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\dfrac{3}{14}\)---------------------->\(\dfrac{3}{14}\)
\(\Rightarrow V_{H_2}=\dfrac{3}{14}.22,4=4,8\left(l\right)\)
b) \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
PTHH: \(ZnO+H_2\xrightarrow[]{t^o}Zn+H_2O\)
Xét tỉ lệ: \(0,1< \dfrac{3}{14}\Rightarrow H_2\) dư
Theo PT: \(n_{Zn}=n_{ZnO}=0,1\left(mol\right)\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(1mol\) \(1mol\)
\(\dfrac{3}{14}mol\) \(\dfrac{3}{14}mol\)
\(a)n_{Fe}=\dfrac{m}{M}=\dfrac{12}{56}\approx0,21=\dfrac{3}{14}\left(mol\right)\)
\(V_{H_2}=n.22,4=\dfrac{3}{14}.22,4=4,8\left(l\right)\)
\(b)n_{ZnO}=\dfrac{m}{M}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
\(ZnO+H_2\rightarrow Zn+H_2O\)
\(1mol\) \(1mol\) \(1mol\)
\(0,1mol\) \(0,1mol\) \(0,1mol\)
\(\text{Ta thấy }H_2\text{ dư,ZnO phản ứng hết.Bài toán tính theo ZnO}\)
\(m_{Zn}=n.M=0,1.65=6,5\left(g\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
a. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Theo PTHH: \(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow m_{muối}=0,1.161=16,1\left(g\right)\)
b. \(n_{H_2thu.được}=n_{Zn}=0,1\left(mol\right)\)
\(H_2+\dfrac{1}{2}O_2\underrightarrow{t^o}H_2O\)
0,1 0,05
\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)
\(\Rightarrow V_{không.khí}=1,12.5=5,6\left(l\right)\)
$a) 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$b) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)$
Theo PTHH : $n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,6(mol)$
$m_{H_2SO_4} = 0,6.98 = 58,8(gam)$
$c) n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol) \Rightarrow m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)$
$d) n_{H_2} = n_{H_2SO_4} = 0,6(mol) \Rightarrow V_{H_2} = 0,6.22,4 = 13,44(lít)$
\(a,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(b.n_{Al}=\dfrac{m}{M}=0,4\left(mol\right)\)
\(Theo.PTHH\Rightarrow n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}n_{Al}=1,5.0,4=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=n.M=0,6.98=58,8\left(g\right)\)
\(c,Theo.PTHH\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,5.0,4=0,2\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=n.M=0,2.342=68,4\left(g\right)\\ d,V_{H_2\left(dktc\right)}=n.22,4=0,6.22,4=13,44\left(l\right)\)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right);n_{ZnCl_2}=\dfrac{27,2}{136}=0,2\left(mol\right)\)
PTHH(1): 2Al + 6HCl → 2AlCl3 + 3H2
Mol: 0,2 0,3
PTHH(2): Zn + 2HCl → ZnCl2 + H2
Mol: 0,2 0,2 0,2
Ta có: \(n_{H_2\left(1\right)}=0,5-0,2=0,3\left(mol\right)\)
\(m_{hh}=0,2.27+0,2.65=18,4\left(g\right)\)
\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.1.................................0.1\)
\(Đặt:n_{CuO\left(pư\right)}=x\left(mol\right)\)
\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)
\(x............x\)
\(m_{cr}=6-80x+64x=5.2\left(g\right)\)
\(\Rightarrow x=0.05\)
\(H\%=\dfrac{0.05}{0.075}\cdot100\%=66.67\%\)
\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.n_{Zn}=n_{ZnCl_2}=0,1\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\ c.n_{H_2}=n_{Zn}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
a) Zn + 2HCl → ZnCl2 + H2
b) mZnCl2 = 0,1 . 136 = 13,6 gam
c) 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.
\(Zn+2HCl->ZnCl_2+H_2\\ m_{Zn}=\dfrac{7,437}{24,79}\cdot65=19,5g\\ m_{HCl}=\dfrac{7,437}{24,79}\cdot2\cdot36,5=21,9g\)