cho 19g hh Na2CO3 và NaHCO3 tác dụng với 100g dd HCl sinh ra 4,48l khí (đktc). a)viết pthh xảy ra b) Tính thành phần % theo khối lượng của mỗi muối trong hh ban đầu
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
tl............1................2.............2.............1.............1..(mol
br 0,1.................0,2......................................0,1(mol)
NaCl không phản ứng đc vsHCl
b)\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(C_{MHCl}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)(đổi 400ml=0,4(l))
c)\(Tacom_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)
\(\Rightarrow\%m_{Na_2CO_3}=\dfrac{10,6}{20}.100=53\%\)
\(\Rightarrow\%mNaCl=100\%-53\%=47\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
448ml = 0,448l
\(n_{CO2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
a) Pt : \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O|\)
1 2 2 1 1
0,02 0,02
b) \(n_{Na2CO3}=\dfrac{0,02.1}{1}=0,02\left(mol\right)\)
\(m_{Na2CO3}=0,02.106=2,12\left(g\right)\)
\(m_{NaCl}=5-2,12=2,88\left(g\right)\)
c) 0/0NaCl = \(\dfrac{2,88.100}{5}=57,6\)0/0
0/0Na2CO3 = \(\dfrac{2,12.100}{5}=42,4\)0/0
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Na2CO3+2HCl=>2NaCl+CO2\(\uparrow\)+H2O
x =>x
0,1mol=>0,2mol
NaHCO3+HCl=>NaCl+CO2\(\uparrow\)+H2O
y =>y
0,1mol=>0,1mol
nCO2=0,2mol
\(|^{106x+84y=19}_{x+y=0,2}\)\(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
mNa2CO3=0,1\(\times\)106=10,6g
%Na2CO3=\(\dfrac{10,6\times100}{19}\)=56%
%NaHCO3=100\(-\)56=44%
nHCL=0,1+0,2=0,3mol
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
x___________________ x______________
0,1_______________0,2______________mol
\(NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O\)
y____________________ y_________________
0,1___________________0,1__________ mol
\(n_{CO2}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}106x+84y=0\\x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(m_{Na2CO3}=0,1.106=10,6\left(g\right)\)
\(m_{NaHCO3}=19-10,6=8,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nH2= 0,35(mol)
a) PTHH: Mg + 2 HCl -> MgCl2 + H2
x_________2x_______x______x(mol)
PTHH: Fe + 2 HCl -> FeCl2 + H2
y________2y________y_____y(mol)
Ta có hpt: \(\left\{{}\begin{matrix}24x+56y=13,2\\x+y=0,35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,15\end{matrix}\right.\)
b) m=m(muối khan)= mMgCl2 + mFeCl2= 95.x+127y=95.0,2+127.0,15= 38,05(g)
a)
Gọi
\(n_{Fe} = a(mol) ; n_{Mg} = b(mol)\\ \Rightarrow 56a + 24b = 13,2(1)\)
\(Mg + 2HCl \to MgCl_2 + H_2\\ Fe + 2HCl \to FeCl_2 + H_2\)
Theo PTHH : \(n_{H_2} = a + b = 0,35(mol)\)(2)
Từ (1)(2) suy ra a = 0,15 ;b = 0,2
Vậy :
\(\%m_{Fe} = \dfrac{0,15.56}{13,2}.100\% = 63,64\%\\ \Rightarrow m_{Mg} = 100\% - 63,64\% = 36,36\%\)
b)
Ta có :\(n_{HCl} = 2n_{H_2} = 0,7(mol)\)
Bảo toàn khối lượng :
\(m_{muối} = m_{kim\ loại} + m_{HCl} - m_{H_2} = 13,2 + 0,7.36,5 - 0,35.2=38,05(gam)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: đktc → đkc
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: 24nMg + 56nFe = 13,2 (1)
\(n_{H_2}=\dfrac{8,6765}{24,79}=0,35\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=0,35\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,2\left(mol\right)\\n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2.24}{13,2}.100\%\approx36,36\%\\\%m_{Fe}\approx63,64\%\end{matrix}\right.\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\)
⇒ m muối khan = 0,2.95 + 0,15.127 = 38,05 (g)
Giải:
a) Số mol khí CO2 sinh ra là:
nCO2 = V/22,4 = 4,48/22,4 = 0,2 (mol)
PTHH: Na2CO3 + 2HCl -> 2NaCl + H2CO3
PTHH: 10NaHCO3 + 10HCl -> 10NaCl + H2O + 15CO2↑
--------------\(\dfrac{2}{15}\)------------------------------------------0,2--
b) Khối lượng NaHCO3 là:
mNaHCO3 = n.M = \(\dfrac{2}{15}\).84 = 11,2 (g)
Thành phần phần trăm theo khối lượng của NaHCO3 trong hỗn hợp ban đầu là:
%mNaHCO3 = (mNaHCO3/mhh).100 = (11,2/19).100 ≃ 58,95 %
=> %mNa2CO3 = 100 - 58,95 = 41,05 %
Vậy ...
\(\text{a) }Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\left(1\right)\\ NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O\left(2\right)\)
\(\text{b) }n_{CO_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\left(1\right)\\ \text{ }\text{ }\text{ }x\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\\ NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O\left(2\right)\\ \text{ }\text{ }\text{ }y\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }y\)
Từ \(\left(1\right)\) và \(\left(2\right),\) ta có hệ phương trình: \(\left\{{}\begin{matrix}x+y=0,2\\106x+84y=19\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\Rightarrow m_{Na_2CO_3}=n\cdot M=0,1\cdot106=10,6\left(g\right)\\ m_{NaHCO_3}=n\cdot M=0,1\cdot84=8,4\left(g\right)\)
\(\Rightarrow\%Na_2CO_3=\dfrac{10,6\cdot100}{19}=55,79\%\\ \%NaHCO_3=\dfrac{8,4\cdot100}{19}=44,21\%\)