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\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\
V_{H_2}=0,1.22,4=2,24l\\
m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2
\(m_{HCl}=0,4.36,5=14,6g\\
V_{H_2}=0,2.22,4=4,48l\\
m\text{dd}=4,8+200-0,4=204,4g\\
C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)
a.b.\(n_{Zn}=\dfrac{1,95}{65}=0,03mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,03 0,06 0,03 0,03 ( mol )
\(V_{H_2}=0,03.22,4=0,672l\)
\(m_{ddHCl}=\dfrac{0,06.36,5}{7,3\%}=30g\)
c.Tên muối: Kẽm clorua
\(m_{ZnCl_2}=0,03.136=4,08g\)
\(m_{ddspứ}=30+1,95-0,03.2=31,89g\)
\(C\%_{ZnCl_2}=\dfrac{4,08}{31,89}.100\%=12,79\%\)
a,\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,15 0,3
Ta có: \(\dfrac{0,2}{1}>\dfrac{0,15}{1}\) ⇒ H2 pứ hết,Fe dư
\(V_{H_2}=3,36\left(l\right)\) (đề cho)
b, ko tính đc k/lg dd ,chỉ tính đc thể tích dd
\(V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)
\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ n_{ZnCl_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ m_{ZnCl_2} = 0,1.136 = 13,6(gam)\\ b) n_{H_2} = n_{Zn} = 0,1(mol) \Rightarrow V_{H_2} = 0,1.22,4 =2 ,24(lít)\\ c) n_{HCl} =2 n_{H_2} = 0,2(mol)\\ \Rightarrow m_{HCl} = 0,2.36,5 = 7,3(gam)\ ; V_{dd\ HCl} = \dfrac{0,2}{0,5} = 0,4(lít)\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
c, \(C_{M_{ZnCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
\(n_{HCl}=0,2.1=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
a, \(n_{Fe}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
b, \(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
c, \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(\text{nH2 = 0,5 mol}\)
PTHH: Fe + 2HCl \(\rightarrow\) FeCl2 +H2
______0,2___0,4_____0,2____0,2___(mol)
\(\text{x = mFe = 0,2. 56 = 11,2 gam}\)
mdd sau phản ứng = mFe + mdd HCl - mH2
\(\text{= 11,2 + 400 - 0,2. 2 }\)
= 410,8 gam
C% FeCl2 = \(\text{(0,2. 127. 100)(410,8)= 6,18%}\)
a) 2Al + 6HCl --> 2AlCl3 + 3H2
b) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\); nHCl = 0,5.2 = 1 (mol)
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{1}{6}\) => Al hết, HCl dư
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,1->0,3---->0,1---->0,15
=> VH2 = 0,15.22,4 = 3,36 (l)
c) \(\left\{{}\begin{matrix}C_{M\left(AlCl_3\right)}=\dfrac{0,1}{0,5}=0,2M\\C_{M\left(HCl.dư\right)}=\dfrac{1-0,3}{0,5}=1,4M\end{matrix}\right.\)
\(n_{H_2}=\dfrac{20,16}{22,4}=0,9\left(mol\right)\)
PTHH :
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,6 1,8 0,6 0,9
\(a,m_{HCl}=1,8.36,5=65,7\left(g\right)\)
\(C\%_{HCl}=\dfrac{65,7}{400}.100\%=16,425\%\)
\(b,m_{AlCl_3}=0,6.133,5=80,1\left(g\right)\)
\(m_{ddAlCl_3}=\left(0,6.27+400\right)-0,9.2=414,4\left(g\right)\)
\(C\%_{AlCl_3}=\dfrac{80,1}{414,4}.100\%\approx19,33\%\)