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)

 NAOH 10© ?

mNaOH=200.10%=20(g)

=>nNaOH=20/40=0,5(mol)

PTHH: NaOH + HCl -> NaCl + H2O

nHCl=nNaOH=0,5(mol) => mHCl=0,5 x 36,5= 18,25(g)

mddHCl=(18,25.100)/3,65= 500(g)

\(m_{NaOH}=\dfrac{200.10}{100}=20\left(g\right)\\ n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

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

\(n_{HCl}=n_{NaOH}=0,5\left(mol\right)\\ m_{HCl}=0,5.36,5=18,25\left(g\right)\\ m_{ddHCl}=\dfrac{18,25.100}{3,65}=500\left(g\right)\)

Trả lời:

mk chx hok wa lớp 9 nên ko giúp đc, thông cảm

HT^^

\(NaOH+HCl->NaCl+H_2O\)

a, \(m_{HCl}=\frac{C\%.m_{\text{dd}HCl}}{100\%}=\frac{7,3\%.200}{100\%}=14.6g\)

\(n_{HCl}=\frac{m_{HCl}}{M_{HCl}}=\frac{14.6}{36.5}=0.4\left(mol\right)\)

Theo PTHH ta có:\(n_{HCl}=n_{NaOH}=0.4\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,4.40=16g\)

\(\Rightarrow m_{\text{dd}NaOH}=\frac{m_{NaOH}.100\%}{C\%}=\frac{16.100\%}{10\%}=160g\)

b, Ta có \(\frac{C\%_{\text{dd}NaOH}-C\%_{\text{dd}mu\text{ối}}}{C\%_{\text{dd}mu\text{ối}}-C\%_{\text{dd}HCl}}=\frac{m_{\text{dd}HCl}}{m_{\text{dd}NaOH}}\)

\(\Leftrightarrow\frac{10\%-C\%}{C\%-7,3\%}=\frac{200}{160}=\frac{5}{4}\)\(\Rightarrow4\left(10\%-C\%\right)=5\left(C\%-7.3\%\right)\Leftrightarrow40\%-4C\%=5C\%-36.5\%\)

\(\Leftrightarrow9C\%=76.5\%\Leftrightarrow C\%=8,5\%\)

Bài trên:

\(n_{NaOH}=\dfrac{16\%.200}{40}=0,8\left(mol\right)\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=\dfrac{0,8}{2}=0,4\left(mol\right)\\ m_{ddH_2SO_4}=\dfrac{0,4.98.100}{19,6}=200\left(g\right)\)

Bài sau:

\(Đặt:\left\{{}\begin{matrix}a=n_{Al}\\b=n_{Mg}\end{matrix}\right.\left(a,b>0\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+24b=12,6\\22,4.1,5a+22,4b=13,44\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\\ \Rightarrow m_{muối}=0,2.133,5+0,3.95=55,2\left(g\right)\\ x=C_{MddHCl}=\dfrac{0,2.3+0,3.2}{0,2}=6\left(M\right)\)

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%

Câu 1:

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

Ta có: \(n_{HCl}=0,2\cdot2=0,4\left(mol\right)=n_{NaOH}\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

Câu 2: Bạn xem lại đề !!

Câu 2:

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

Ta có: \(n_{NaOH}=\dfrac{200\cdot10\%}{40}=0,5\left(mol\right)=n_{HCl}\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5\cdot36,5}{3,65\%}=500\left(g\right)\)

Câu 3:

\(NaHCO_3+HCl\rightarrow NaCl+H_2O+CO_2\uparrow\)

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

\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

*Các chất trên tác dụng tương tự với H₂SO₄ loãng 

Bonus:  \(Ba\left(NO_3\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HNO_3\)

a)

$n_{HCl} = \dfrac{200.10\%}{36,5} = \dfrac{40}{73}(mol)$
$n_{CaCO_3} = \dfrac{10}{100} = 0,1(mol)$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{CaCO_3} : 1 < n_{HCl} : 2$ nên HCl dư
$n_{CO_2} = n_{CaCO_3} = 0,1(mol) \Rightarrow V_{CO_2} = 0,1.22,4 = 2,24(lít)$
b)

$n_{HCl\ dư} = \dfrac{40}{73} - 0,1.2 = 0,348(mol)$
$NaOH + HCl \to NaCl + H_2O$
$n_{NaOH} = n_{HCl} = 0,348(mol)$
$C_{M_{NaOH}} = \dfrac{0,348}{1} = 0,348M$