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a ,\(n_{CO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
b, \(PTHH:CaCO_3+2HCl-->CaCl_2+H_2O+CO_2\)
\(BaCO_3+2HCl-->BaCl_2+H_2O+CO_2\)
c, Đặt số mol \(CaCO_3:a\left(mol\right);n_{BaCO_3}=b\left(mol\right)\)
Ta có hệ : \(100a+197b=69,7\)
\(a+b=0,6\)
\(=>\left\{{}\begin{matrix}a=0,5\\b=0,1\end{matrix}\right.\)
\(=>\%CaCO_3=\dfrac{0,5.100}{69,7}=71,74\%\)
\(=>\%BaCO_3=...\)
d, \(n_{HCl}=2n_{CO_2}=2.0,6=1,2\left(mol\right)\)
\(=>m_{HCl}=43,8\left(g\right)\)
\(=>m_{\text{dd}HCl_{7,3\%}}=\dfrac{100}{7,3}.43,8=600\left(g\right)\)
2al+ 6hcl-> 2alcl3+3h2
a-> 3a a 1,5a
fe+2hcl-> fecl2+h2
b->2b b b
27a+56b= 5,5
1,5a+b=4,48/22,4
=> a=0,1; b=0,05
=> %mal=0,1*27/5,5*100=49,09%
=>%mfe= 100-49,09=50,9%
mhcl= 3a+2b= 3*0,1+2*0,05=0,4
=>mddhcl= 0,4*36,5*100/14,6=100g
-> vddhcl=100/ 1,08=92,592ml
mddsau pư= 5,5+100-0,2*2=105,1
C% alcl3= 133,5*0,1/105,1*100=12,7
Cfecl2= 127* 0,05/105,1*100=6,04
Na2CO3(x) + 2HCl ---> 2NaCl + CO2(x ) +H2O
K2CO3(y ) + 2HCl ---> 2KCl + CO2(y ) +H2O
Đặt nNa2CO3 = x (mol); nK2CO3 = y (mol)
=> 106x + 138y = 38,2 (1)
nCO2 = 0,3 (mol)
=> x + y = 0,3 (mol) (2)
Từ (1,2) => x = 0,1 (mol) , y = 0,2 (mol)
=> % khối lượng
Theo PTHH: nHCl = 2nCO2 = 0,6 (mol)
\(\Rightarrow m_{ddHCl}=\dfrac{0,6.36,5.100}{10}=219\left(g\right)\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=0,1(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,1.24}{6,4}.100\%=37,5\%\\ \Rightarrow \%_{MgO}=100\%-37,5\%=62,5\%\)
\(b,n_{MgO}=\dfrac{6,4-0,1.24}{40}=0,1(mol)\\ \Rightarrow n_{HCl}=2.0,1+2.0,1=0,4(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,4}{0,5}=0,8(l)\\ c,n_{MgCl_2}=0,1+0,1=0,2(mol)\\ \Rightarrow C_{M_{MgCl_2}}=\dfrac{0,2}{0,8}=0,25M\)
\(a,Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ b,n_{Fe}=n_{H_2}=0,2\left(mol\right)\\ \%m_{Fe}=\dfrac{0,2.56}{12,8}.100\%=87,5\%\\ \%m_{Fe_2O_3}=100\%-87,5\%=12,5\%\\ c,n_{Fe_2O_3}=\dfrac{12,8-11,2}{160}=0,01\left(mol\right)\\ n_{H_2SO_4}=n_{Fe}+3n_{Fe_2O_3}=0,2+3.0,01=0,23\left(mol\right)\\ V_{ddH_2SO_4}=\dfrac{0,23}{0,46}=0,5\left(M\right)\)
1.1. Al + NaOH + H2O ==> NaAlO2 + 3/2H2
nH2(1)=3,36/22,4=0.15(mol)
=> nAl(1)= nH2(1):3/2= 0.15:3/2= 0.1(mol)
2.Mg + 2HCl ==> MgCl2 + H2
3.2Al + 6HCl ==> 2AlCl3 + 3H2
4.Fe + 2HCl ==> FeCl2 + H2
=> \(n_{H_2\left(2,3,4\right)}=\) 10.08/22.4= 0.45(mol)
=> nH2(3)=0.1*3/2=0.15(mol)
MgCl2 + 2NaOH ==> Mg(OH)2 + 2NaCl
AlCl3 + 3NaOH ==> Al(OH)3 + 3NaCl
FeCl2 + 2NaOH ==> Fe(OH)2 + 2NaCl
BaCO3 + H2SO4 -> BaSO4 + CO2 + H2O (1)
CaCO3 + H2SO4 -> CaSO4 + CO2 + H2O (1)
nCO2=0,3(mol)
Đặt nBaCO3=a
nCaCO3=b
Ta có:
\(\left\{{}\begin{matrix}197a+100b=39,7\\a+b=0,3\end{matrix}\right.\)
=>a=0,1;b=0,2
mBaCO3=197.0,1=19,7(g)
mCaCO3=100.0,2=20(g)
%mBaCO3=\(\dfrac{19,7}{39,7}.100\%=49,622\%\)
%mCaCO3=100-49,622=50,378%
d;
Theo PTHH 1 và 2 ta có:
nBaCO3=nH2SO4(1)=0,1(mol)
nCaCO3=nH2SO4(2)=0,2(mol)
mH2SO4=98.0,3=29,4(g)
mdd =29,4:20%=147(g)