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-nNa2CO3= m/M = 10,6/106 = 0,1 (mol)
-PT:Na2CO3+CaCl2->CaCO3+2NaCl
____0,1____________0,1______0,2
-mCaCO3= n.M = 0,1.100 = 10 (g)
-mNaCl= n.M = 0,2.58,5 = 11,7 (g)
\(n_{CaCl_2}=\dfrac{22.2}{111}=0.2\left(mol\right)\)
\(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\)
\(0.2....................................................0.4\)
\(m_{AgCl}=0.4\cdot143.5=57.4\left(g\right)\)
nCaCl2=22,2/111=0,2(mol)
CaCl2 + 2AgNO3 -----> 2AgCl + Ca(NO3)2
TPT:nAgCl=2.nCaCl2=2.0,2=0,4(mol)
mAgCl=0,4.143,5=57,4(g)
Gọi KL cần tìm là M
\(n_{AgNO_3}=\dfrac{170}{170}=1(mol)\\ MCl_2+2AgNO_3\to M(NO_3)_2+2AgCl\downarrow\\ \Rightarrow n_{MCl_2}=\dfrac{1}{2}n_{AgNO_3}=0,5(mol)\\ \Rightarrow M_{MCl_2}=\dfrac{55,5}{0,5}=111(g/mol)\\ \Rightarrow M_M=111-35,5.2=40(g/mol)(Ca)\\ n_{Ca(NO_3)_2}=0,5(mol);n_{AgCl}=1(mol)\\ \Rightarrow m_{Ca(NO_3)_2}=0,5.164=82(g);m_{AgCl}=1.143,5=143,5(g)\)
\(a,n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Ba + 2H2O ---> Ba(OH)2 + H2
0,3<-------------0,3<---------0,3
=> mBa = 0,3.137 = 41,1 (g)
=> mK2O = 59,9 - 41,1 = 18,8 (g)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{41,1}{59,9}.100\%=68,61\%\\\%m_{K_2O}=100\%-68,61\%=31,39\%\end{matrix}\right.\)
\(b,n_{K_2O}=\dfrac{18,8}{94}=0,2\left(mol\right)\)
PTHH: K2O + H2O ---> 2KOH
0,2----------------->0,4
Các chất tan trong dd sau phản ứng: KOH, Ba(OH)2
\(\rightarrow\left\{{}\begin{matrix}m_{KOH}=0,4.56=22,4\left(g\right)\\m_{Ba\left(OH\right)_2}=0,3.171=51,3\left(g\right)\end{matrix}\right.\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
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
\(m_{HCl}=0,4.36,5=14,6g\\
V_{H_2}=0,2.22,4=4,48l\\
m\text{dd}=4,8+200-0,4=204,4g\\
C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)
Pt:
2\(Fe+3CaCl_2\)➞3\(Ca+2FeCl_3\)
2 3 3 2
0,2 0,3 0,3 0,2
\(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
➞ \(m_{Ca}=n.M=0,3.40=12\left(g\right)\)
➞\(m_{FeCl_3}=n.M=0,2.162,5=32,5\left(g\right)\)
Zn+S->ZnS
0,2-------0,2
n Zn=\(\dfrac{13}{65}\)=0,2 mol
n S=\(\dfrac{9,6}{32}\)=0,3 mol
=>S dư
=>m S=0,1.32=3,2g
=>m ZnS=0,2.97=19,4g
\(PTHH:CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\downarrow\)
\(n_{CaCl_2}=\frac{22,2}{111}=0,2\left(mol\right)\)
Theo PT: \(n_{AgNO_3}=2n_{CaCl_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{AgNO_3}=0,4.170=68\left(g\right)\)
b) Các chất còn lại trong phản ứng là Ca(NO3)2, AgCl
\(TheoPT:n_{Ca\left(NO_3\right)_2}=n_{CaCl_2}=0,2\left(mol\right)\)
\(n_{AgCl}=2n_{CaCl_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{Ca\left(NO_3\right)_2}=0,2.164=32,8\left(g\right)\)
\(m_{AgCl}=0,4.143,5=57,4\left(g\right)\)
CaCl2 + 2AgNO3 → Ca(NO3)2 + 2AgCl
\(n_{CaCl_2}=\frac{22,2}{111}=0,2\left(mol\right)\)
a) Theo PT: \(n_{AgNO_3}=2n_{CaCl_2}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{AgNO_3}=0,4\times170=68\left(g\right)\)
b) Theo PT: \(n_{Ca\left(NO_3\right)_2}=n_{CaCl_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Ca\left(NO_3\right)_2}=0,2\times164=32,8\left(g\right)\)
Theo PT: \(n_{AgCl}=n_{AgNO_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{AgCl}=0,4\times143,5=57,4\left(g\right)\)