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\(PTHH:CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\downarrow\)
\(n_{CaCl_2}=\frac{22,2}{111}=0,2\left(mol\right)\)
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: \(n_{AgCl}=n_{AgNO_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{AgCl}=0,4\times143,5=57,4\left(g\right)\)
\(n_{CaCl_2}=\dfrac{22.2}{111}=0.2\left(mol\right)\)
\(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\)
\(0.2....................................................0.4\)
\(m_{AgCl}=0.4\cdot143.5=57.4\left(g\right)\)
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)
-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)
Gọi KL cần tìm là M
\(n_{AgNO_3}=\dfrac{170}{170}=1(mol)\\ MCl_2+2AgNO_3\to M(NO_3)_2+2AgCl\downarrow\\ \Rightarrow n_{MCl_2}=\dfrac{1}{2}n_{AgNO_3}=0,5(mol)\\ \Rightarrow M_{MCl_2}=\dfrac{55,5}{0,5}=111(g/mol)\\ \Rightarrow M_M=111-35,5.2=40(g/mol)(Ca)\\ n_{Ca(NO_3)_2}=0,5(mol);n_{AgCl}=1(mol)\\ \Rightarrow m_{Ca(NO_3)_2}=0,5.164=82(g);m_{AgCl}=1.143,5=143,5(g)\)
Bài 1 :
\(a) CaCO_3 \xrightarrow{t^o} CaO + CO_2\\ 2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O\\ b) CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O\\ 2Fe(OH)_3 + 6HCl \to 2FeCl_3 + 6H_2O\\ NaOH + HCl \to NaCl + H_2O\\ c) 2AgNO_3 + 2NaOH \to 2NaNO_3 + Ag_2O + H_2O\\ NaCl + AgNO_3 \to AgCl + NaNO_3\)
a, PT: \(CaCl_2+2AgNO_3\rightarrow2AgCl_{\downarrow}+Ca\left(NO_3\right)_2\)
b, Ta có: \(n_{CaCl_2}=\dfrac{2,22}{111}=0,02\left(mol\right)\)
\(n_{AgNO_3}=\dfrac{1,7}{170}=0,01\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,02}{1}>\dfrac{0,01}{2}\), ta được CaCl2 dư.
Theo PT: \(n_{AgCl}=n_{AgNO_3}=0,01\left(mol\right)\Rightarrow m_{AgCl}=0,01.143,5=1,435\left(g\right)\)
c, \(n_{CaCl_2\left(pư\right)}=\dfrac{1}{2}n_{AgNO_3}=0,005\left(mol\right)\)
\(\Rightarrow n_{CaCl_2\left(dư\right)}=0,015\left(mol\right)\Rightarrow m_{CaCl_2\left(dư\right)}=0,015.111=1,665\left(g\right)\)
8.5g AgNO3 hay 85g??
nMgCl2 = 19/95=0.2mol
nAgNO3=8.5/170=0.05mol
MgCl2 + 2AgNO3 -> Mg(NO3)2 + 2AgCl
(mol) 1 2
(mol) 0.2 0.05
Lập tỉ lệ: 0.2/1>0.05/2. vậy MgCl2 dư
MgCl2 + 2AgNO3 -> Mg(NO3)2 + 2AgCl
(mol) 0.025 0.05 0.025 0.05
mMg(NO3)2 = 0.025*148=3.7g
mAgCl = 0.05*143.5=7.175g
8.5g AgNO3 hay 85g??
\(a,n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
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
\(\rightarrow\left\{{}\begin{matrix}m_{KOH}=0,4.56=22,4\left(g\right)\\m_{Ba\left(OH\right)_2}=0,3.171=51,3\left(g\right)\end{matrix}\right.\)
nCACL2=0.2mol
CaCl2+2AgNO3->CA(NO3)2+2AGCL
->nAgNO3=0,4 mol
mAgNo3=68g
mCa(NO3)2=32,8
mAgCl=28,7g