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Bài 9:
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{14,6\%.100}{36,5}=0,4\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,4}{2}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,4-2.0,1=0,2\left(mol\right)\\ m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4 0,2 ( mol )
\(m_{HCl}=0,4.36,5=14,6g\)
\(m_{ddHCl}=\dfrac{14,6\times100}{14,6}=100g\)
\(m_{ddspứ}=100+13=113g\)
\(m_{ZnCl_2}=0,2.136=27,2g\)
\(C\%_{ZnCl_2}=\dfrac{27,2}{113}.100=24,07\%\)
Bài 14 :
\(a) n_{CuO} = \dfrac{8}{80} = 0,1(mol)\\ CuO + 2HCl \to CuCl_2 + H_2O\\ n_{HCl} = 2n_{CuO} = 0,2(mol)\\ m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)\\ b) \text{Chất tan : } CuCl_2\\ n_{CuCl_2} = n_{CuO} = 0,1(mol)\\ m_{CuCl_2} = 0,1.135 = 13,5(gam)\)
Bài 15 :
\(a) n_{Fe_2O_3} =\dfrac{4,8}{160} = 0,03(mol)\\ Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O\\ n_{H_2SO_4} = 3n_{Fe_2O_3} = 0,09(mol)\\ m_{dd\ H_2SO_4} = \dfrac{0,09.98}{9,8\%} = 90(gam)\\ b) \text{Chất tan : } Fe_2(SO_4)_3\\ n_{Fe_2(SO_4)_3} = n_{Fe_2O_3} = 0,03(mol)\\ m_{Fe_2(SO_4)_3} = 0,03.400 = 12(gam)\)
\(n_{HCl}=\dfrac{80\cdot14.6\%}{36.5}=0.32\left(mol\right)\)
\(n_{FeO}=\dfrac{7.2}{72}=0.1\left(mol\right)\)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\)
Lập tỉ lệ :
\(\dfrac{0.1}{1}< \dfrac{0.32}{2}\Rightarrow HCldư\)
\(m_{HCl\left(dư\right)}=\left(0.32-0.1\cdot2\right)\cdot36.5=4.38\left(g\right)\)
\(m_{FeCl_2}=0.1\cdot127=12.7\left(g\right)\)
\(m_{HCl}=\dfrac{14,6.80}{100}=11,68\left(g\right)\Rightarrow n_{HCl}=\dfrac{11,68}{36,5}=0,32\left(mol\right)\);
\(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)
\(PTHH:FeO+2HCl\rightarrow FeCl_2+H_2O\)
Mol: 0,1 0,2 0,1
Ta có tỉ lệ:\(\dfrac{0,32}{2}>\dfrac{0,1}{1}\) =>HCl dư,FeO phản ứng hết
a)mHCl dư=(0,32-0,2).36,5=4,38 (g)
b)\(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
a, nH2=5,6/22,4=0,25 mol
Zn+2HCl->ZnCl2+H2
0,25 0,5 0,25 0,25
mZn pư=0,25.65=16,25 g
b, C%HCl=0,5.36,5.100/200=9,125%
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\
pthh:Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\\
m_{HCl}=\dfrac{\left(0,2.36,5\right).100}{24,5}=29,795\left(g\right)\\
m_{\text{dd}}=5,6+29,795-\left(0,1.2\right)=35,195\left(g\right)\\
C\%=\dfrac{0,1.127}{35,195}.100\%=36\%\)
Theo gt ta có: $n_{Zn}=0,06(mol)$
$Zn+2HCl\rightarrow ZnCl_2+H_2$
a, Ta có: $n_{HCl}=0,12(mol)\Rightarrow m_{ddHCl}=30(g)$
b, Ta có: $n_{H_2}=0,06(mol)\Rightarrow V_{H_2}=1,344(l)$
\(_{ }\)nHCL =\(\dfrac{3,65}{100x36,5}\)x103=1 mol
nCAO=0,1 mol
CAO +2HCL→CACL2+ H2O
0,1---> 0,2
=> HCL DƯ
a)
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{Al} = \dfrac{10,8}{27} = 0,4(mol)$
Theo PTHH : $n_{HCl} = 3n_{Al} = 1,2(mol)$
$\Rightarrow m = \dfrac{1,2.36,5}{14,6\%} = 300(gam)$
b) $n_{H_2} = \dfrac{3}{2}n_{Al} = 0,6(mol)$
$M_xO_y + yH_2 \xrightarrow{t^o}xM + yH_2O$
Theo PTHH : $n_{oxit} = \dfrac{1}{y}.n_{H_2} = \dfrac{0,6}{y}(mol)$
$\Rightarrow \dfrac{0,6}{y}(Mx + 16y) = 34,8$
$\Rightarrow \dfrac{x}{y}.M = 42$
Với x = 3 ; y = 4 thì $M = 56(Fe)$
Vậy oxit là $Fe_3O_4$
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
\(n_{CaO}=0,1mol\)
\(\rightarrow n_{HCl}=2n_{CaO}=0,2mol\)
\(\rightarrow m_{HCl}=7,3g\)
\(\rightarrow m_{HCldd}=\frac{7,3.100}{14,6}=50g\)
\(\rightarrow m_{dd}=m_{CaO}+m_{HCldd}=5,6+50=55,6g\)