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Câu 3 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,6.100}{10,95}=133,3\left(g\right)\)
c) \(n_{H2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
d) \(n_{MgCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{MgCl2}=0,2.95=19\left(g\right)\)
\(m_{ddspu}=4,8+133,3-\left(0,2.2\right)=137,7\left(g\right)\)
\(C_{MgCl2}=\dfrac{19.100}{137,7}=13,8\)0/0
Chúc bạn học tốt
nH2=\(\frac{6,72}{22,4}=0,3\)mol
PTHH
M+2HCl--> MCl2+H2
0,3mol<---------------0,3mol
=>MM=\(\frac{19,5}{0,3}=64\)
=> km loại là kẽm (Zn)
b) nNaOH=0,2.1=0,2 mol
PTHH
NaOH+HCl-->NaCl + H2O
0,2 mol--> 0,2 mol
---> thể tích HCl 1M đã dùng là V=\(\frac{0,2+0,3}{1}=0,5\)lít
=> CM(ZnCl2)=\(\frac{0,3}{0,5}=0,6M\)
200ml = 0,2l
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,3 0,6 0,3 0,3
a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,6 0,6
\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)
\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)
Chúc bạn học tốt
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
______0,2---->0,3------------>0,1------>0,3______(mol)
=> VH2 = 0,3.22,4= 6,72(l)
b) \(C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,1}=3M\)
\(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,1}{0,1}=1M\)
Câu 3:
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=n_{H_2}=0,3(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,3.24}{15,2}.100\%=47,37\%\\ \Rightarrow \%_{MgO}=100\%-47,37\%=52,63\%\)
\(n_{MgO}=\dfrac{15,2-0,3.24}{40}=0,2(mol)\\ \Rightarrow \Sigma n_{HCl}=0,3.2+0,2.2=1(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{1.36,5}{10\%}=365(g)\\ \Sigma n_{MgCl_2}=0,2+0,3=0,5(mol)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{0,5.95}{15,2+365}.100\%=12,49\%\)
\(PTHH:Mg+2H_2SO_{4(đ)}\to MgSO_4+2H_2O+SO_2\uparrow\\ MgO+H_2SO_4\to MgSO_4+H_2O\\ \Rightarrow n_{SO_2}=n_{Mg}=0,3(mol)\\ \Rightarrow V_{SO_2}=0,3.22,4=6,72(l)\)
a+b) PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)
Ta có: \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{CO_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\\V_{CO_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=0,1\left(mol\right)\\n_{NaOH}=\dfrac{50\cdot40\%}{40}=0,5\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\) Tạo muối trung hòa, bazơ dư, tính theo CO2
Bảo toàn Cacbon: \(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\) \(\Rightarrow m_{Na_2CO_3}=0,1\cdot106=10,6\left(g\right)\)
Bài 7:
Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\n_{NaOH}=0,2\cdot1=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối
PTHH: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
a_______2a__________a (mol)
\(CO_2+NaOH\rightarrow NaHCO_3\)
b_______b__________b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}a+b=0,15\\2a+b=0,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{CO_2}+m_{ddNaOH}=0,15\cdot44+200\cdot1,25=256,6\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2CO_3}=\dfrac{0,05\cdot106}{256,6}\cdot100\%\approx2,1\%\\C\%_{NaHCO_3}=\dfrac{0,1\cdot72}{256,6}\cdot100\%\approx2,8\%\end{matrix}\right.\)
Bài 8:
PTHH: \(RCO_3+2HNO_3\rightarrow R\left(NO_3\right)_2+CO_2\uparrow+H_2O\)
Giả sử \(n_{RCO_3}=1\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{HNO_3}=2\left(mol\right)\\n_{R\left(NO_3\right)_2}=1\left(mol\right)=n_{CO_2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ddHNO_3}=\dfrac{2\cdot63}{20\%}=630\left(g\right)\\m_{R\left(NO_3\right)_2}=R+124\left(g\right)\\m_{CO_2}=44\left(g\right)\end{matrix}\right.\) \(\Rightarrow C\%_{R\left(NO_3\right)_2}=\dfrac{124+R}{R+60+630-44}=0,26582\)
\(\Leftrightarrow R=65\) (Kẽm) \(\Rightarrow\) CTHH của muối cacbonat là ZnCO3
1. nNaOH = 300. 20% : 40 = 1,5 mol
Pư: 3NaOH + H3PO4 ----> Na3PO4 + 3H2O
1,5 mol -----0,5 mol
=> mH3PO4 = 0,5. 98 = 49g
2, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(n_{HCl}=\dfrac{200}{1000}.2,5=0,5\left(mol\right)\)
2Al + 6HCl ----> 2AlCl3 + 3H2
Ta co: \(\dfrac{0,1}{2}< \dfrac{0,5}{6}\Rightarrow\) HCl dư
2Al + 6HCl ----> 2AlCl3 + 3H2
de: 0,1.......0,5
pu: 0,1......0,3................0,1.........0,15
spu: 0 .......0,2.................0,1........0,15
a, \(V_{H_2}=0,15.22,4=3,36l\)
b, \(m_{HCl}=1,25.200=250g\)
\(m_{dd}=250+2,7-0,15.2=252,4g\)
\(m_{HCl\left(dư\right)}=0,2.36,5=7,3g\)
\(m_{AlCl_3}=0,1.133,5=13,35g\)
\(C\%_{HCl\left(dư\right)}=\dfrac{7,3}{252,4}.100\%\approx2,89\%\)
\(C\%_{AlCl_3}=\dfrac{13,35}{252,4}.100\%\approx5,29\%\)