Ta có: \(m_{NaOH}=200.20\%=40\left(g\right)\Rightarrow n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

a, Theo PT: \(n_{HCl}=n_{NaOH}=1\left(mol\right)\)

\(\Rightarrow C\%_{HCl}=\dfrac{0,1.36,5}{100}.100\%=36,5\%\)

b, \(n_{NaCl}=n_{NaOH}=1\left(mol\right)\)

Ta có: m _{dd sau pư} = 200 + 100 = 300 (g)

\(\Rightarrow C\%_{NaCl}=\dfrac{1.58,5}{300}.100\%=19,5\%\)

a. Ta có: m_NaOH=\(\frac{200.20}{100}=40\left(g\right)\)

pt :         NaOH               +              HCl            -------->           NaCl         +        H₂O

theo pt:    40g                                  36,5g                               58,5g                  18g

theo đề:  40g                                   36,5g                               58,5g

=>\(C_{\%}=\frac{58,5}{200+100}.100\%=19,5\%\) 

b.\(C_{\%}=\frac{36,5}{100}.100\%=36,5\%\)

https://hoc24.vn/cau-hoi/cho-400-g-dung-dich-naoh-30-tac-dung-vua-het-voi-100-g-dung-dich-hcl-tinh-a-nong-do-muoi-thu-duoc-sau-phan-ungb-tinh-nong-do-axit-hcl-bie.7974814552205

Tui trả lời rùi nghen

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1           0,2            0,1      0,1 
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\ V_{H_2}=0,1.22,4=2,24l\\ m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)  
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2             0,4       0,2              0,2 
\(m_{HCl}=0,4.36,5=14,6g\\ V_{H_2}=0,2.22,4=4,48l\\ m\text{dd}=4,8+200-0,4=204,4g\\ C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)

\(m_{NaOH}=200\cdot20\%=40\left(g\right)\)

\(\Rightarrow n_{NaOH}=0,1\left(mol\right)\)

m_{dd ( sau pư)} = 200 + 100 = 300 (g)

NaOH+HCl => NaCl+H₂O

1mol=>1mol=>1mol

\(\Rightarrow C\%_{dd}NaCl=\frac{58,5}{100}\times100\%=19,5\%\)

\(m_{HCl}=36,5\left(g\right)\)

\(\Rightarrow C\%_{dd}HCl=\frac{36,5}{100}\cdot100\%=36,5\%\)

Bài 1

PTHH : NaOH + HCl -> NaCl + H₂O

a, m_{dung dịch} = 200 + 100 = 300 (g)

C%_HCl= 100/300 .100% =33,33%

b, m_NaOH=200.20%=40(g)

=> n_{NaOH =}1(mol)

Theo PTHH , n_NaCl = n_NaOH = 1(mol)

=> m_NaCl=58,5 (g)

=> C% = 58,5/300 . 100% = 19,5%

Bài 2

PTHH : Mg + H₂SO4 -> MgSO₄ + H₂

m_{H2SO4=29,4(g)}=> n_H2SO4 = 29,4/98=0,3(mol)

m_Mg= 2,4/24=0,1 (mol)

Vì ),1<0,3=> Mg phản ứng hết , tính theo số mol Mg

Theo PTHH , n_H2=n_Mg = 0,1 (mol)=>V_H2(dktc) = 2,24 (l)

b, Theo PTHH , n_H2SO4= 0,1 (mol)=> n_{H2SO4 dư} = 0,2 (mol)=> m_{H2SO4 dư} = 0,2 .98 = 19,6 (g)

Theo PTHH , n_MgSO4 = 0,1 (mol)=> m_MgSO4 = 0,1 . (24+32+16.4)=12(g)

=> m_{dung_dịch _ sau _Phản _ứng}= 12+19,6 =31,6 (g)

C%_MgSO4= 12/31,6 . 100% \(\approx38\%\)

C%_{H2SO4 dư} = 62%

a) PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

b) Ta có: \(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,2\cdot95=19\left(g\right)\\C\%_{HCl}=\dfrac{0,4\cdot36,5}{200}\cdot100\%=7,3\%\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{MgO}+m_{ddHCl}=208\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{19}{208}\cdot100\%\approx9,13\%\)

c) PTHH: \(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_2\downarrow\)

Ta có: \(\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{NaOH}=\dfrac{200\cdot4\%}{40}=0,2\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) \(\Rightarrow\) NaOH p/ứ hết, MgCl₂ còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,2\left(mol\right)\\n_{Mg\left(OH\right)_2}=0,1\left(mol\right)=n_{MgCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{MgCl_2\left(dư\right)}=9,5\left(g\right)\\m_{Mg\left(OH\right)_2}=0,1\cdot58=5,8\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{ddA}+m_{ddNaOH}-m_{Mg\left(OH\right)_2}=402,2\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{402,2}\cdot100\%\approx2,91\%\\C\%_{MgCl_2\left(dư\right)}=\dfrac{9,5}{402,2}\cdot100\%\approx2,36\%\end{matrix}\right.\) 

Theo gt ta có: $n_{MgO}=0,2(mol)$

a, $MgO+2HCl\rightarrow MgCl_2+H_2O$

b, Ta có: $n_{HCl}=0,4(mol)\Rightarrow x=7,3$

Bảo toàn khối lượng ta có: $m_{ddA}=208(g)$

$\Rightarrow \%C_{MgCl_2}=9,13\%$

c, Ta có: $n_{NaOH}=0,2(mol)$

$\Rightarrow n_{Mg(OH)_2}=0,1(mol)$

Bảo toàn khối lượng ta có: $m_{ddB}=208+200-0,1.58=402,2(g)$

$\Rightarrow \%C_{MgCl_2}=2,36\%$

mNaOH=200.20%=40g

=>nNaOH=1 mol

$NaOH$ + $HCl$ => $NaCl$ + $H_2O$

1 mol                        =>1 mol

=>C% dd sau pứ=58,5/(200+100).100%=19,5%

m$HCl$ =36,5g

=>C% dd HCl=36,5/100.100%=36,5%

Câu 2

100ml=0,1(l)

PTHH : Fe + 2HCl -> FeCl₂ + H₂

n_Fe = 5,6/56=0,1(mol)

n_HCl = 0,1 (mol)

Vi 0,1<0,1/2=> Fe dư , HCl hết , tính theo n_HCl

Theo PTHH , n_H2= 0,05(mol)=> V_H2=1,12(l)

b, Theo PTHH , n_{Fe đẵ phản ứng}=0,05(mol)

=>n_{Fe dư}=0,05(mol)=>m=2,8(g)

Khối lượng các chất sau phản ứng = m_Fe + m_FeCl2 =...

sau đó bạn tự tính nhé ......vốn dĩ mik ko định kamf câu này nhưng thấy máy bạn làm sai nên làm

Câu 2:

a) nFe=5,6:56=0,1(mol)

PTHH: Fe + 2HCl -> FeCl2 + H2

theo phương trình ta có: nH2=nFe=0,1(mol)

-> VH2=0,1.22,4=2,24(l)