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a , ta có
pt1: HCl + Na2SO3 --> H2O + 2NaCl + SO2
0,01(mol) 0,01(mol)
pt2 : H2SO4 + Na2SO3 -->H2O + Na2SO2 + SO2
0,02(mol) 0,02(mol)
pt3 : HCl + BaCl2 --> k tác dụng :)
pt4 : H2SO4 + BaCl2 --> 2HCl + BaSO4
0,02(mol) 0,02(mol)
b, nSO2 =\(\frac{0,672}{22,4}\)=0,03( mol) ; nBaSO4=\(\frac{4,66}{233}\)=0,02 ( mol) ; 50 (ml) = 0,05 (lít)
=> nH2SO4=nBaSO4
=0,02 (mol)
từ pt2 ta có : nSO2 = nH2SO4
= 0,02 (mol)
từ pt1 và pt2 ta có : nSO2(pt1)= nSO2 - nSO2(pt2)
=0,03 -0,02 =0,01 (mol)
=> CM HCl = \(\frac{0,01}{0,05}\)= 0,2(M)
=> CM H2SO4=\(\frac{0,02}{0,05}\)=0,4(M)
ko biết đúng ko . :v
Bài 1:
PTHH: 2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2\(\uparrow\)
Theo pt: 2 .......... 3 ................. 1 ............ 3 ... (mol)
Theo đề: 0,2 ..... 0,3 .............. 0,1 ........ 0,3 ... (mol)
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(V_{ddH_2SO_4}=\dfrac{n}{C_M}=\dfrac{0,3}{2}=0,15\left(l\right)=150\left(ml\right)\)
\(V_{H_{2\left(đktc\right)}}=n.22,4=0,3.22,4=6,72\left(l\right)\)
Vậy \(V=150ml\) và \(V'=6,72l\)
Bài 3:
PTHH: MgO + 2HCl \(\rightarrow\) MgCl2 + H2O (1)
Theo pt: . 1 ........ 2 ............. 1 .......... 1 ...... (mol)
Theo đề: 0,12 .. 0,24 .................................. (mol)
PTHH: Fe2O3 + 6HCl \(\rightarrow\) 2FeCl3 + H2O (2)
Theo pt: .. 1 ......... 6 ............ 2 ........... 1 ...... (mol)
Theo đề: 0,12 .... 0,72 .................................. (mol)
Gọi nMgO = nFe2O3 = x (mol)
Ta có: mMgO + mFe2O3 = 24 (gt)
\(\Leftrightarrow\) x(40 + 160) = 24 \(\Leftrightarrow\) 200x = 24 \(\Leftrightarrow\) x = 0,12 (mol)
Lại có: nHCl = nHCl(1) + nHCl(2) = 0,24 + 0,72 = 0,96 (mol)
Do đó: \(V_{ddHCl}=\dfrac{n}{C_M}=\dfrac{0,96}{1,6}=0,6\left(l\right)=600\left(ml\right)\)
Vậy \(V=600ml\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\) (1)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\) (2)
a) Ta có: \(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\) \(\Rightarrow n_{Mg}=0,1mol\)
\(\Rightarrow m_{Mg}=0,1\cdot24=2,4\left(g\right)\) \(\Rightarrow m_{MgO}=4\left(g\right)\)
b) Ta có: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{Mg}=0,2mol\\n_{HCl\left(2\right)}=2n_{MgO}=2\cdot\frac{4}{40}=0,2mol_{ }\end{matrix}\right.\)
\(\Rightarrow n_{HCl}=0,4mol\) \(\Rightarrow V_{ddHCl}=\frac{0,4}{1}=0,4\left(l\right)=400\left(ml\right)\)
c) Ta có: \(\left\{{}\begin{matrix}n_{MgCl_2\left(1\right)}=n_{Mg}=0,1mol\\n_{MgCl_2\left(2\right)}=n_{MgO}=0,1mol\end{matrix}\right.\)
\(\Rightarrow m_{muối}=m_{MgCl_2\left(1\right)}+m_{MgCl_2\left(2\right)}=0,1\cdot40+0,1\cdot40=8\left(g\right)\)
Mg +2HCl\(\rightarrow\)MgCl2+ H2
MgO+2HCl\(\rightarrow\)MgCl2+H2O
ta có nH2 =\(\frac{2,24}{22,4}\)=0,1 mol
Theo pthh 1 :nH2=nMg =0,1 mol
\(\rightarrow\)mMg=0,1.24 =2,4 g
\(\rightarrow\)mMgO =6-2,4=3,6g
nMgO =\(\frac{3,6}{40}\)=0,09 mol
%mMgO =\(\frac{3,6}{6}.100\%\)=60%
Ta co nHCl =2nMg =0,2 mol
nHCl =2nMgO =0,18 mol
mHCl=(0,18+0,2).36,5=13,87 g
VHCl=\(\frac{13,87}{1}.1\)=12,6ml
Theo đề bài ta có :
m(kết tủa) = mBaSO4 = 17,475 (g) => nBaSO4 = \(\dfrac{17,475}{233}=0,075\left(mol\right)\)
V(khí) = VCO2 => nCO2 = \(\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Ta có PTHH :
\(\left(1\right)CaCO3+H2SO4\rightarrow C\text{aS}O4+H2O+CO2\)
0,05mol.............0,05mol.......0,05mol
\(\left(2\right)BaCl2+H2SO4\rightarrow B\text{aS}O4\downarrow+2HCl\)
0,075mol............0,075mol.....0,075mol...0,15mol
a) Ta có : nH2SO4 = 0,05 + 0,075 = 0,125 (mol)
=> mddH2SO4 = \(\dfrac{0,0125.98}{9,8}.100=125\left(g\right)\)
=> VddH2SO4=\(\dfrac{m}{D}=\dfrac{12,5}{1,12}\approx111,61\left(ml\right)\)
b) DD Y bao gồm CaSO4 và HCl
=> \(\left\{{}\begin{matrix}C\%_{C\text{aS}O4}=\dfrac{0,05.136}{0,05.100+125-0,05.44}.100\%\approx5,32\%\\C\%_{HCl}=\dfrac{0,15.36,5}{0,075.208+125-17,475}.100\%\approx4,45\%\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}CM_{C\text{aS}O4}=\dfrac{0,05}{0,11161}\approx0,45\left(M\right)\\CM_{HCl}=\dfrac{0,15}{0,11161}\approx1,34\left(M\right)\end{matrix}\right.\)
Vậy..............