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\(m_{HCl}=200.14,6\%=29,2\left(g\right)\)
\(n_{HCl}=\frac{29,2}{36,5}=0,8\left(mol\right)\)
\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
x__________________2x______x____________
\(KHCO_3+HCl\rightarrow KCl+CO_2+H_2O\)
y__________________y______y____________
\(n_{NaOH}=0,4.1=0,4\left(mol\right)\)
Giải hệ PT:
\(\left\{{}\begin{matrix}106x+100y=25,9\\2x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
\(\Rightarrow n_{CO2}=x=0,15\left(mol\right)\)
a,\(V_{CO2}=0,15.22,4=3,36\left(l\right)\)
b,
\(\%_{NaCl}=\frac{\left(2.0,15\right).58,5}{\left(2.0,15\right).58,5+0,1.74,5}.100\%=70,2\%\)
\(\Rightarrow\%_{KCl}=100\%-70,2\%=29,8\%\)
Ta có: \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{HCl}=0,4.1,5=0,6\left(mol\right)\)
Giả sử: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=x\left(mol\right)\\n_{K_2CO_3}=y\left(mol\right)\end{matrix}\right.\)
PT: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
\(K_2CO_3+2HCl\rightarrow2KCl+H_2O+CO_2\)
\(HCl_{dư}+NaOH\rightarrow NaCl+H_2O\)
Theo PT: \(n_{CO_2}=n_{Na_2CO_3}+n_{K_2CO_3}=x+y\left(mol\right)\) ⇒ x + y = 0,25 (1)
\(n_{HCl\left(pư\right)}=2x+2y\left(mol\right)\) \(\Rightarrow n_{HCl\left(dư\right)}=0,6-2x-2y\left(mol\right)\)
Có: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}+n_{HCl\left(dư\right)}=0,6-2y\left(mol\right)\\n_{KCl}=2n_{K_2CO_3}=2y\left(mol\right)\end{matrix}\right.\)
⇒ 58,5(0,6 - 2y) + 74,5.2y = 39,9 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{0,1.106}{0,1.106+0,15.138}.100\%\approx33,9\%\\\%m_{K_2CO_3}\approx66,1\%\end{matrix}\right.\)
Bạn tham khảo nhé!
Gọi số mol CO2 và SO2 là a, b (mol)
= >\(\left\{{}\begin{matrix}n_{khí}=a+b=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\\dfrac{44a+64b}{a+b}=29,5.2=59\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,3\left(mol\right)\end{matrix}\right.\)
\(n_{NaOH}=1.0,4=0,4\left(mol\right)\)
PTHH: NaOH + CO2 --> NaHCO3
________0,1<----0,1------->0,1_______(mol)
NaOH + SO2 --> NaHSO3
_0,3<----0,3-------->0,3_____________(mol)
=> \(\left\{{}\begin{matrix}C_{M\left(NaHCO_3\right)}=\dfrac{0,1}{0,4}=0,25M\\C_{M\left(NaHSO_3\right)}=\dfrac{0,3}{0,4}=0,75M\end{matrix}\right.\)
nH2= 2.24: 22.4 = 0,1(mol)
nNaAlO2= 16,4:82=0,2(mol)
Gọi nAl=a (mol;a>0)
nNaAlO2= b (mol; b>0)
(1) Al + NaOH + H2O --> NaAlO2 + 3/2 H2
Ta có a a a 1,5a (mol)
(2) Al2O3 + 2NaOH --> 2NaAlO2 + H2O
Ta có b 2b (mol)
Ta có nH2 = 0,1 (mol) => 1,5a = 0,1 => a = 2/30=>nAl = 2/30(mol)
nNaAlO2 = 0,2(mol) =>a+2b=0,2=> b=??
Còn lại bn tự làm nha!!!!!!!!
Có chỗ nào sai mong mọi người thông cảm:))
\(n_{CO2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH : \(CaO+2HCl\rightarrow CaCl_2+H_2O\) (1)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\) (2)
a) Theo Pt : \(n_{CO2}=n_{CaCO3}=0,1\left(mol\right)\)
\(m_{CaCO3}=0,1100=10\left(g\right)\)
\(m_{CaO}=12,8-10=2,8\left(g\right)\)
b) Chắc tính V của dd HCl đã dùng
(1) \(n_{CaO}=\dfrac{2,8}{56}=0,05\left(mol\right)\) , \(n_{HCl}=2n_{CaO}=0,1\left(mol\right)\)
(2) \(n_{HCl}=2n_{CaCO3}=0,2\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{0,1+0,2}{1}=0,3\left(l\right)=300\left(ml\right)\)
\(n_{H_2}=\dfrac{3,36}{22,4}0,15(mol)\\ a,PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ b,n_{Fe}=n_{H_2}=0,15(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,15.56}{14,8}.100\%=56,76\%\\ \Rightarrow \%_{Cu}=100\%-56,76\%=43,24\%\\ c,n_{H_2SO_4}=0,15(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,15.98}{20\%}=73,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{73,5}{1,4}=52,5(l)\)
a. \(m_{HCl}=200.14,6\%=29,2g\)
\(n_{HCl}=\frac{29,2}{36,5}=0,8mol\)
\(n_{NaOH}=\frac{400}{1000}.1=0,4mol\)
Để trung hoà hết Y cần 0,4mol NaOH, suy ra HCl dư
\(n_{HCl_{pu}}=0,8-0,4=0,4mol\)
PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\)
\(KHCO_3+HCl\rightarrow KCl+CO_2\uparrow+H_2O\)
Đặt \(\hept{\begin{cases}x\left(mol\right)=n_{Na_2CO_3}\\y\left(mol\right)=n_{KHCO_3}\end{cases}}\)
\(\rightarrow106x+100y=25,9g\left(1\right)\)
Theo phương trình \(n_{HClpu}=2n_{Na_2CO_3}+n_{KHCO_3}=0,4mol\)
\(\rightarrow2x+y=0,4mol\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}x=0,15mol\\y=0,1mol\end{cases}}\)
Theo phương trình \(n_{CO_2}=n_{Na_2CO_3}+n_{KHCO_3}=0,15+0,1=0,25mol\)
\(V_{CO_2}=0,25.22,4=5,6l\)
b. \(m_{dd}=25,9+200-0,25.44=214,9g\)
\(C\%_{NaCl}=\frac{2.0,15.58,5}{214,9}.100\%=8,2\%\)
\(C\%_{KCl}=\frac{0,1.74,5}{214,9}.100\%=3,5\%\)