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a) CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
b) \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
_0,1---->0,2------->0,1----->0,1
=> mCaCl2 = 0,1.111 = 11,1 (g)
=> VCO2 = 0,1.22,4 = 2,24 (l)
c) \(a=C_{M\left(HCl\right)}=\dfrac{0,2}{0,4}=0,5M\)
d) \(C_{M\left(CaCl_2\right)}=\dfrac{0,1}{0,4}=0,25M\)
a) Na2SO3 + 2HCl --> 2NaCl + SO2 + H2O
b) nHCl = 0,2.1 = 0,2 (mol)
Na2SO3 + 2HCl --> 2NaCl + SO2 + H2O
_0,1<------0,2------->0,2----->0,1
mNaCl = 0,2.58,5 = 11,7(g)
VSO2 = 0,1.22,4 = 2,24 (l)
c) mNa2SO3 = 0,1.126 = 12,6 (g)
d) \(C_{M\left(NaCl\right)}=\dfrac{0,2}{0,2}=1M\)
\(n_{MnO_2}=\dfrac{69,6}{87}=0,8\left(mol\right)\)
nKOH = 0,5.4 = 2(mol)
PTHH: MnO2 + 4HCl --> MnCl2 + Cl2 + 2H2O
0,8------------------------>0,8
2KOH + Cl2 --> KCl + KClO + H2O
Xét tỉ lệ \(\dfrac{2}{2}>\dfrac{0,8}{1}\) => KOH dư, Cl2 hết
2KOH + Cl2 --> KCl + KClO + H2O
1,6<--0,8---->0,8---->0,8
=> \(\left\{{}\begin{matrix}n_{KOH\left(dư\right)}=2-1,6=0,4\left(mol\right)\\n_{KCl}=0,8\left(mol\right)\\n_{KClO}=0,8\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C_{M\left(KOH\right)}=\dfrac{0,4}{0,5}=0,8M\\C_{M\left(KCl\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(KClO\right)}=\dfrac{0,8}{0,5}=1,6M\end{matrix}\right.\)
Đáp án C
= 0,8 mol
MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
0,8 → 
= 0,72 (mol)
Vkhí = 0,72.22,4 = 16,128 (lit)
nNaOH = 2 (mol)
Cl2 + 2NaOH → NaCl + NaClO + H2O
0,72 2 → 0,72 0,72 (mol)
do NaOH dư, tính theo Cl2
Dung dịch sau phản ứng: nNaCl = nNaClO = 0,72 (mol)
nNaOH dư = 0,56 (mol)
CNaCl = CNaClO = 1,44M, CNaOH = 1,12M
\(a,Fe+2HCl\rightarrow FeCl_2+H_2\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\%\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
a) PTHH: MnO2 + 4 HCl -> MnCl2 + Cl2 + 2 H2O
0,12___________0,48______0,12___0,12(mol)
nMnO2 = 10,44/ 87=0,12(mol)
V(Cl2, đktc)= 0,12. 22,4= 2,688(l)
b) NaOH gì vậy em??
a) PTHH: MnO2 + 4 HCl -> MnCl2 + Cl2 + 2 H2O
0,12___________0,48______0,12___0,12(mol)
nMnO2 = 10,44/ 87=0,12(mol)
V(Cl2, đktc)= 0,12. 22,4= 2,688(l)
Cl2+2NaOH→NaCl+NaClO
⇒nCl2=nMnO2=0,12mol
⇒nNaOH=2nCl2=0,24mol
⇒VNaOH=0,12M
⇒{CM(NaCl)=0,06M
CM(NaClO)=0,06M