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\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
0,1 0,1
\(m_{Fe}=0,1.56=5,6\left(g\right)\\ \%m_{Fe}=\dfrac{5,6}{12,8}=43,75\%\\ \%m_{FeO}=100\%-43,75\%=56,25\%\)
a, Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{MgCO_3}=b\left(mol\right)\end{matrix}\right.\)
\(n_{hhkhí\left(H_2,CO_2\right)}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH:
Zn + H2SO4 ---> ZnSO4 + H2
a a a
MgCO3 + H2SO4 ---> MgSO4 + CO2 + H2O
b b b
Hệ pt \(\left\{{}\begin{matrix}a+b=0,2\\161a+84b=28,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Zn}=0,1.65=6,5\left(g\right)\\m_{MgCO_3}=0,1.84-8,4\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{6,5}{6,5+8,4}=43,62\%\\\%m_{MgCO_3}=100\%-43,62\%=56,38\%\end{matrix}\right.\)
b, \(n_{H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\)
PTHH:
2Na + H2SO4 ---> Na2SO4 + H2
0,03 0,015 0,015
\(\rightarrow m_{Al_2\left(SO_4\right)_3}=7,26-0,015.142=5,13\left(g\right)\\ \rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{5,13}{342}=0,015\left(mol\right)\)
Al2O3 + 3H2SO4 ---> Al2(SO4)3 + 3H2O
0,015 0,015
\(\rightarrow\left\{{}\begin{matrix}m_{Na}=0,03.23=0,69\left(g\right)\\m_{Al_2O_3}=0,015.102=1,53\left(g\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,69}{0,69+1,53}=31,08\%\\\%m_{Al_2O_3}=100\%-31,08\%=68,92\%\end{matrix}\right.\)
c, Thiếu \(d_{H_2SO_4}\)
nHCl=0,7.1=0,7(mol)
nCO2=4,48/22,4=0,2(mol)
PTHH: MgCO3 +2 HCl -> MgCl2 + CO2 + H2O
0,2_________0,4_____0,2______0,2(mol)
nMgCO3=nCO2=0,2(mol) => mMgCO3=0,2. 84= 16,8(g)
=> mFeO= mX - mMgCO3= 24 - 16,8= 7,2(g)
=> %mMgCO3= (16,8/24).100=70%
=>%mFeO=100% - 70%= 30%
b) nFeO= 7,2/72=0,1(mol)
FeO +2 HCl -> FeCl2 + H2O
0,1___0,2______0,1(mol)
Vì 0,4+0,2=0,6 => HCl có dư => nHCl(dư)= 0,7 - 0,6=0,1(mol)
Vddsau= VddHCl=0,7(l)
CMddHCl(dư)= 0,1/0,7= 1/7 (M)
CMddFeCl2= 0,1/0,7=1/7(M)
CMddMgCl2= 0,2/0,7=2/7(M)
a. \(n_{CO_2}=0,2\left(mol\right)\)
\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)
\(\Rightarrow n_{MgCl_2}=n_{MgCO_3}=n_{CO_2}=0.2\left(mol\right)\)
\(\Rightarrow m_{MgCO_3}=0,2.84=16,8\left(g\right)\)
\(\%m_{MgCO_3}=\dfrac{16,8.100\%}{24}=70\%\\ \%m_{FeO}=100\%-70\%=30\%\)
b. \(FeO+2HCl\rightarrow FeCl_2+H_2O\)
\(n_{FeO}=\dfrac{24-16,8}{72}=0,1\left(mol\right)=n_{FeCl_2}\)
\(C_{M_{FeCl_2}}=\dfrac{0,1}{0,7}=\dfrac{1}{7}\left(M\right)\)
\(C_{M_{MgCO_3}}=\dfrac{0,2}{0,7}=\dfrac{2}{7}\left(M\right)\)
a) \(\left\{{}\begin{matrix}160n_{Fe_2O_3}+80n_{CuO}=24\\n_{Fe_2O_3}=n_{CuO}\end{matrix}\right.\Rightarrow n_{Fe_2O_3}=n_{CuO}=0,1\)
\(\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{160.0,1}{24}.100\%=66,67\%\\\%m_{CuO}=\dfrac{80.0,1}{24}.100\%=33,33\%\end{matrix}\right.\)
PTHH: Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
0,1------>0,3-------->0,1
CuO + H2SO4 --> CuSO4 + H2O
0,1-->0,1---------->0,1
nCuSO4 = 0,1 (mol)
nFe2(SO4)3 = 0,1 (mol)
=> m = 0,1.160 + 0,1.400 = 56(g)
b) \(m_{H_2SO_4\left(pthh\right)}=\left(0,3+0,1\right).98=39,2\left(g\right)\)
=> mH2SO4(thực tế) = \(\dfrac{39,2.125}{100}=49\left(g\right)\)
c) \(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
=> nBaSO4 = 0,5 (mol)
=> mBaSO4 = 0,5.233 = 116,5(g)
Bài 1:
TN1: moxit = 78,4 (g)
Gọi số mol H2SO4 pư là a (mol)
=> \(n_{H_2O}=a\left(mol\right)\) (Bảo toàn H)
Theo ĐLBTKL: \(m_{oxit}+m_{H_2SO_4}=m_{muối}+m_{H_2O}\)
=> 78,4 + 98a = 190,4 + 18a
=> a = 1,4 (mol)
=> \(n_O=n_{H_2O}=1,4\left(mol\right)\)
=> mkim loại = moxit - mO = 78,4 - 1,4.16 = 56 (g)
TN2:
Gọi \(\left\{{}\begin{matrix}n_{HCl}=x\left(mol\right)\\n_{H_2SO_4}=y\left(mol\right)\end{matrix}\right.\)
Bảo toàn H: \(n_{H_2O}=\dfrac{x+2y}{2}=1,4\left(mol\right)\)
=> x + 2y = 2,8 (1)
Có: \(m_{muối}=m_{kim.loai}+m_{Cl}+m_{SO_4}\)
=> 175,4 = 56 + 35,5x + 96y
=> 35,5x + 96y = 119,4 (2)
(1)(2) => x = 1,2; y = 0,8
Vậy số mol H2SO4 trong Y là 0,8 (mol)
Bài 2:
Gọi số mol Fe trong 15,8 gam X là a (mol)
TN1:
Bảo toàn Fe: \(n_{FeCl_2}=a\left(mol\right)\)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
=> nHCl = 1,2 (mol)
Theo ĐLBTKL: mkim loại + mHCl = mmuối + mH2
=> mmuối = 15,8 + 1,2.36,5 - 0,6.2 = 58,4 (g)
TN2:
Bảo toàn Fe: \(n_{FeCl_3}=a\left(mol\right)\)
Khối lượng muối tăng: 61,95 - 58,4 = 3,55 (g)
=> \(m_{FeCl_3}-m_{FeCl_2}=3,55\)
=> 162,5a - 127a = 3,55
=> a = 0,1 (mol)
\(\%m_{Fe}=\dfrac{0,1.56}{15,8}.100\%=35,443\%\)
a) \(2Fe+6H_2SO_4\text{đặc}\rightarrow^{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
\(2FeO+4H_2SO_4\text{đặc}\rightarrow^{t^0}Fe_2\left(SO_4\right)_3+SO_2+4H_2O\)
\(2Fe_3O_4+10H_2SO_4\text{đặc}\rightarrow3Fe_2\left(SO_4\right)_3+SO_2+10H_2O\)
Quy đổi hỗn hợp A gồm x mol Fe và y mol O
\(\Rightarrow56x+16y=49,6\) (1)
\(Fe\rightarrow Fe^{3+}+3e\)
x ------------> 3x
\(S^{+6}+2e\rightarrow S^{+4}\)
..........0,8 <--- 0,4
\(O+2e\rightarrow O^{2-}\)
y --> 2y
\(\Rightarrow3x=2y+0,8\) (2)
Giải hệ (1)(2) được x = 0,7 mol, y = 0,65 mol
\(\Rightarrow\%m_O=\dfrac{16.0,65}{49,6}.100\%=20,97\%\)
Ta có:
\(2H_2SO_4+2e\rightarrow SO_4^{2-}+SO_2\uparrow+2H_2O\)
...............................0,4 <---- 0,4
\(\Rightarrow n_{SO_4^{2-}}\text{tạo muối với cation kim loại}=0,4\) mol
\(\Rightarrow m_{\text{muối}}=m_{\text{kim loại}}+m_{SO_4^{2-}}\text{tạo muối với cation kim loại}\)
\(=56.0,7+96.0,4=77,6\) gam