nOH-=0,05.2.0,2+ 0,1.0,2=0,04 mol

nH+=0,1.V.2+ 0,2.V=0,4V mol

H++OH--->H2O

nH+=nOH-=> 0,4v=0,04-> v=0,1 lit = 100ml

\(n_{HCl}=0,2.2=0,04\left(mol\right)\)

Pt: \(NaOH+HCl\rightarrow NaCl+H_2O\)

0,04mol \(\leftarrow\) 0,04mol \(\rightarrow\) 0,04mol

\(V_{NaOH}=\dfrac{0,04}{0,1}=0,4\left(l\right)\)

\(\Sigma_{V\left(spu\right)}=0,2+0,4=0,6\left(l\right)\)

\(C_{M_{NaCl}}=\dfrac{0,04}{0,6}=0,067M\)

b) Pt: \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

0,02mol \(\leftarrow\) 0,04mol\(\rightarrow\) 0,02mol

\(m_{Ca\left(OH\right)_2}=0,02.74=1,48\left(g\right)\)

\(m_{dd}=\dfrac{1,48.100}{5}=29,6\left(g\right)\)

\(C\%_{CaCl_2}=\dfrac{0,02.111.100}{29,6}=7,5\%\)

n_HCl=C_M.V=1.0,6=0,6(mol)

Pt: HCl+ NaOH-> NaCl + H₂O

cứ:1.............1.............1 (mol)

vậy:0,6----->0,6------>0,6(mol)

=> m_NaOH=0,6.40=24(g)

\(\Rightarrow m_{ddNaOH}=\dfrac{m_{NaOH}.100\%}{C\%}=\dfrac{24.100}{30}=80\left(g\right)\)

b) Nếu thay NaOH bằng Ca(OH)₂ thì ta có PT:

PT: 2HCl + Ca(OH)₂ -> CaCl₂ +2H₂O

Cứ:2................1.................1 (mol)

Vậy: 0,6------->0,3----------->0,3(mol)

=> m_Ca(OH)2=n.M=0,3.74=22,2(g)

\(\Rightarrow m_{ddCa\left(OH\right)_2}=\dfrac{m_{Ca\left(OH\right)_2}.100\%}{C\%}=\dfrac{22,2.100}{7,351}\approx302\left(g\right)\)

\(\Rightarrow V_{ddCa\left(OH\right)_2}=\dfrac{m_{ddCa\left(OH\right)_2}}{D}=\dfrac{302}{1,045}\approx289\left(ml\right)=0,289\left(lít\right)\)

Ta có : \(n_{OH}=2C_{MBa\left(OH\right)2}V+C_{MKOH}V=2,09\left(mol\right)\)

\(BTNT\left(O\right):n_{H2O}=n_{OH}=2,09\left(mol\right)\)

\(BTNT\left(H\right):n_H=n_{H2O}=2,09\left(mol\right)\)

Mà \(n_H=2,09=1,98V+2,2V=4,18V\)

\(\Rightarrow V=0,5\left(l\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Ba\left(OH\right)2}=0,76\\n_{H2SO4}=0,55\end{matrix}\right.\) ( mol )

\(PTHH:H_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4+2H_2O\)

Thấy ; \(0,55< 0,76\)

\(\Rightarrow n_{BaSO4}=0,55\left(mol\right)\)

\(\Rightarrow m_{kt}=128,15g\)

1 ) nOH= nNaOH+2nBa(OH)2=0,01+2.0,01=0,03mol 
OH- + H+ ----------->H2O 
0.03-->0.03 
nHCl=nH+ =0.03mol 
=>VHCl=0.03/0.3=0.1l=100ml

2 )  (NH4)2SO4 + 2KOH -(t°)-> K2SO4 + 2NH3 + 2H2O 
0,05______________________________ 0,1 (mol) 
n(NH4)2SO4 = 0,05.1 = 0,05 mol 
V NH3 = 0,1 . 22,4 = 2,24 l 

HCl tinh bang 2M

\(n_{Ca\left(OH\right)_2}=0,3.1=0,3\left(mol\right)\\ n_{HCl}=0,2.0,2=0,04\left(mol\right)\)

a

\(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

0,02<------0,04----->0,02

Xét \(\dfrac{0,3}{1}>\dfrac{0,04}{2}\Rightarrow Ca\left(OH\right)_2.dư\)

\(m_{CaCl_2}=0,02.111=2,22\left(g\right)\)

b

Muốn pứ xảy ra hoàn toàn phải thêm dung dịch HCl 0,2 M

\(n_{HCl.cần}=2n_{Ca\left(OH\right)_2}=0,3.2=0,6\left(mol\right)\\ n_{HCl.cần.thêm}=0,6-0,04=0,56\left(mol\right)\)

\(V_{cần.\left(HCl\right)}=\dfrac{0,56}{0,2}=2,8\left(l\right)=280\left(ml\right)\\ V_{cần.thêm\left(HCl\right)}=280-200=80\left(ml\right)\)

c

\(CM_{CaCl_2}=\dfrac{0,02}{0,3+0,28}=\dfrac{1}{29}M\)

\(n_{Ca\left(OH\right)_2}=0,3.1=0,3\left(mol\right)\\ n_{HCl}=0,2.0,2=0,04\left(mol\right)\\ Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

  0,02<----0,04------>0,02

a

\(m_{CaCl_2}=0,02.111=2,22\left(g\right)\)

b

Muốn pứ xảy ra hoàn toàn thì phải thêm dd HCl 0,2M.

\(n_{HCl.cần}=0,3.2=0,6\left(mol\right)\\ n_{HCl.cần.thêm}=0,6-0,04=0,56\left(mol\right)\\ V_{HCl.cần.}=\dfrac{0,56}{0,2}=2,8\left(l\right)\\ V_{HCl.cần.thêm}=2,8-0,2=2,6\left(l\right)=260\left(ml\right)\)

c

\(CM_{CaCl_2}=\dfrac{0,3}{0,3+2,8}=\dfrac{3}{31}M\)

a) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Gọi số mol Zn, Al là a, b

=> 65a + 27b = 18,4 (1)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           a----->2a------------->a

            2Al + 6HCl --> 2AlCl₃ + 3H₂

            b---->3b------------->1,5b

=> a + 1,5b = 0,5 (2)

(1)(2) => a = 0,2 ; b = 0,2

=> \(\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

b) n_HCl(pư) = 2a + 3b = 1 (mol)

n_HCl(dư) = 0,6.2 - 1 = 0,2 (mol)

PTHH: KOH + HCl --> KCl + H₂O

           0,2<----0,2

=> \(V=\dfrac{0,2}{1}=0,2\left(l\right)\)

\(n_{HCl}=0,6.2=1,2\left(mol\right)\\ n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ a,n_{HCl\left(dư\right)}=1,2-2.n_{H_2}=1,2-2.0,5=0,2\left(mol\right)\\PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Đặt:n_{Zn}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}65a+27b=18,4\\a+1,5b=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\end{matrix}\right.\\ \Rightarrow m_{Zn}=0,2.65=13\left(g\right);m_{Al}=0,2.27=5,4\left(g\right)\\ b,KOH+HCl_{dư}\rightarrow KCl+H_2O\\ n_{KOH}=n_{HCl\left(dư\right)}=0,2\left(mol\right)\\ \Rightarrow V=V_{ddKOH}=\dfrac{0,2}{1}=0,2\left(l\right)\)

a)\(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)

\(n_{HCl}=V\cdot C_M=2\cdot0,6=1,2mol\)

\(n_{\uparrow}=\dfrac{11,2}{22,4}=0,5mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

 x         2x           x            x

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

 y         3y           y               1,5y

Ta có hệ: \(\left\{{}\begin{matrix}65x+27y=18,4\\x+1,5y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)

\(m_{Zn}=0,2\cdot65=13g\)

\(m_{Al}=0,2\cdot27=5,4g\)

b)Theo pt: \(\Sigma n_{HCl}=2x+3y=2\cdot0,2+3\cdot0,2=1mol\)

Để trung hòa: \(n_{OH^-}=n_{H^+}=1mol\)

\(\Rightarrow V=\dfrac{n_{OH^-}}{C_M}=\dfrac{1}{1}=1l\)