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\(Zn+2HCl->ZnCl_2+H_2\\ Mg+2HCl->MgCl_2+H_2\\ n_{Zn}=a\\ n_{Mg}=b\\ 65a+24b=11,3g\\ n_{H_2}=a+b=\dfrac{6,72}{22,4}=0,3\\ a=0,1\\ m_{Zn}=65.0,1=6,5g\)
\(a.CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ n_{CaCO_3}=n_{CO_2}=\dfrac{448:1000}{22,4}=0,02\left(mol\right)\\ n_{HCl}=0,02.2=0,04\left(mol\right)\\ C\%_{ddHCl}=\dfrac{0,04.36,5}{1,18.200}\approx0,619\%\\b.m_{CaCO_3}=0,02.100=2\left(g\right)\\ \%m_{CaCO_3}=\dfrac{2}{5}.100=40\%\\ \%m_{CaSO_4}=100\%-40\%=60\% \)
Mình tra KLR của dd HCl trên mạng là 1,18g/ml nên áp dụng vào bài nha ^^
PTHH :
\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)
x 2x 2x x x
\(K_2CO_3+2HCl\rightarrow2KCl+H_2O+CO_2\uparrow\)
y 2y 2y y y
Ta có :
106x + 138y = 26
2x + 2y = 0,4
Giải hệ PT, ta có :
\(\rightarrow x=0,05\left(mol\right);y=0,15\left(mol\right)\)
Thu đc khí CO2 chứ bạn nhỉ?
\(a,V_{CO_2}=\left(0,05+0,15\right).22,4=4,48\left(l\right)\)
\(b,m_{muối}=0,05.58,5+0,15.74,5=14,1\left(g\right)\)
\(c,\%m_{Na_2CO_3}=\dfrac{0,05.106}{26}.100\%\approx20,38\%\)
\(\%m_{K_2CO_3}=100\%-20,38\%=79,62\%\)
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)
c, \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\Rightarrow m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)
d, \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Có: \(\dfrac{0,2}{1}>\dfrac{0,45}{3}\) → Fe2O3 dư.
\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
a) mCu = 1,875 (g)
=> \(\%Cu=\dfrac{1,875}{10}.100\%=18,75\%\)
\(\%Zn=\dfrac{10-1,875}{10}.100\%=81,25\%\)
b) \(m_{Zn}=10-1,875=8,125\left(g\right)\)
=> \(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,125------------------>0,125
=> VH2 = 0,125.22,4 = 2,8 (l)
a)
Mg + 2HCl --> MgCl2 + H2
b)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,3<-----------------0,3
=> mMg = 0,3.24 = 7,2 (g)
=> mAg = 10,4 - 7,2 = 3,2 (g)
c) \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{7,2}{10,4}.100\%=69,23\%\\\%m_{Ag}=\dfrac{3,2}{10,4}.100\%=30,77\%\end{matrix}\right.\)
a, Cu không tác dụng với dd HCl.
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
\(\Rightarrow m_{Cu}=19,4-13=6,4\left(g\right)\)
c, Ta có: \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{19,4}.100\%\approx67,01\%\\\%m_{Cu}\approx32,99\%\end{matrix}\right.\)
Bạn tham khảo nhé!
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
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
\(m_{HCl}=0,4.36,5=14,6g\\
V_{H_2}=0,2.22,4=4,48l\\
m\text{dd}=4,8+200-0,4=204,4g\\
C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)
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: \(\left\{{}\begin{matrix}x+y=0,2\\106x+84y=19\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\Rightarrow m_{Na_2CO_3}=n\cdot M=0,1\cdot106=10,6\left(g\right)\\ m_{NaHCO_3}=n\cdot M=0,1\cdot84=8,4\left(g\right)\)
\(\Rightarrow\%Na_2CO_3=\dfrac{10,6\cdot100}{19}=55,79\%\\ \%NaHCO_3=\dfrac{8,4\cdot100}{19}=44,21\%\)