mHCl=7,3%.200=14,6(g) -> nHCl=0,4(mol)

a) PTHH: HCl + NaOH -> NaCl + H2O

Ta có: nNaOH=nNaCl=nHCl=0,4(mol)

=> mNaOH=0,4.40=16(g)

=>mddNaOH=16:10%=160(g)

=>a=160(g)

b) PTHH: Ba(OH)2 +2 HCl -> BaCl2 + 2 H2O

nBa(OH)2= nHCl/2=0,4/2=0,2(mol)

=> mBa(OH)2=0,2.171=34,2(g)

=>mddBa(OH)2= 34,2 : 17,1%=200(g)

=> Nếu thay dd NaOH bằng dung dịch Ba(OH)2 17,1% thì cần 200 gam dung dịch Ba(OH)2 em nhé!

Đề thiếu rồi bạn

\(m_{Ba\left(OH\right)_2}=\dfrac{342\cdot20\%}{100\%}=68,4\left(g\right)\\ n_{Ba\left(OH\right)_2}=\dfrac{68,4}{137+17\cdot2}=0,4\left(mol\right)\\ PTHH:Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo đề: \(n_{HCl}=2n_{Ba\left(OH\right)_2}=0,8\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,8\cdot36,5=29,2\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{29,2}{100}\cdot100\%=29,2\%\)

\(m_{ct}=\dfrac{20.342}{100}=68,4\left(g\right)\)

\(n_{Ba\left(OH\right)2}=\dfrac{68,4}{171}=0,4\left(mol\right)\)

Pt : \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O|\)

            1                2              1           2

          0,4             0,8

\(n_{HCl}=\dfrac{0,4.2}{1}=0,8\left(mol\right)\)

⇒ \(m_{HCl}=0,8.36,5=29,2\left(g\right)\)

\(C_{ddHCl}=\dfrac{29,2.100}{100}=29,2\)⁰/₀

 Chúc bạn học tốt

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

\(n_{Ba\left(OH\right)_2}=\dfrac{342.20\%}{171}=0,4\left(mol\right)\\ n_{HCl}=2n_{Ba\left(OH\right)_2}=0,8\left(mol\right)\\ C\%_{HCl}=\dfrac{0,8.36,5}{100}.100=29,2\%\)

\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

a, \(n_{H_2}=n_{Zn}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(n_{HCl}=2n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)

c, \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

b, \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{20\%}==73\left(g\right)\)

c, \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,2\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,2}{2,5}=0,08\left(l\right)\)

\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

a, \(n_{H_2}=n_{Mg}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(n_{HCl}=2n_{Mg}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)

c, PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)

nNaOH=200x10/100x40=0.5(mol)

NaOH+HCl-->NaCl+H2O

0.5------0.5                         (mol)

=>mHCl=0.5x36.5=18.25(g)

=>mddHCl=18.25x100/3.65=500(g)

đặt nHCl=a(mol)
=>mHCl=36,5a(g)=>mddHCl=1000a(g)
mNaOH=200.10:100=20g=>nNaOH=0,5(mol)
PTHH:
HCl+NaOH-->NaCl+H2O
0,5__0,5____0,5
theo pt=>md_dHCl=500g
=>m_{ddsau pu}=500+200=700g
tho pt,ta có:n_NaCl=0,5 mol=>m_NaCl=0,5.58,5=29,25g
=>C%(NaCl)=(29,25.100%):700=4,2%

\(n_{HCl} = 0,2.0,2 = 0,04(mol)\\ Ca(OH)_2 + 2HCl \to CaCl_2 + 2H_2O\\ n_{Ca(OH)_2} = \dfrac{n_{HCl}}{2} = 0,02(mol)\\ m_{Ca(OH)_2} = 0,02.74 = 1,48(gam)\\ m_{dung\ dịch\ Ca(OH)_2} = \dfrac{1,48}{5\%} = 29,6(gam)\)

nHCl=0,2.0,2=0,04(mol)

Ca(OH)2+2HCl→CaCl2+2H2O

nCa(OH)2=nHCl2=0,02(mol)

mCa(OH)2= 0,02. 74 =1,48(g)

mddCa(OH)2=1,48.5%=29,6(g)