Bài 42: Nồng độ dung dịch

`n_[CaCl_2]=[55,5]/[111]=0,5(mol)`

`=>C_[M_[CaCl_2]]=[0,5]/[0,5]=1(M)`

\(m_{\text{dd}}=\dfrac{100.0,5}{2,5}=20g\\ m_{H_2O}=20-0,5=19,5g\)

\(V_{\text{dd}}=0,2+0,3=0,5\left(l\right)\\ n_{HCl}=0,2.1+0,3.0,5=0,35\\ C_M=\dfrac{0,35}{0,5}=0,7M\)

200ml = 0,2(l)

=> nHCl (1) = 0,2 .1 = 0,2 (mol)

300ml = 0,3 (l)

=> nHCl(2) = 0,3 . 0,5 = 0,15 (mol)

=> CM (sau khi trộn) = n/V = (0,15+0,2) / (0,2+0,3 ) = 0,35 / 0,5 = 0,7 M

\(PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\)

\(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)

Theo pthh:

\(\left\{{}\begin{matrix}n_{H_2}=\dfrac{1}{2}n_{Na}=1\left(mol\right)\\n_{NaOH}=n_{Na}=2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{H_2}=1.2=2\left(g\right)\\m_{NaOH}=2.40=80\left(g\right)\end{matrix}\right.\)

Theo đlbtkl, ta có:

\(m_{ddNaOH}=m_{Na}+m_{H_2O}-m_{H_2}\)

\(=46+180-2=224\left(g\right)\)

\(C\%ddNaOH=\dfrac{80}{224}.100\approx35,7\%\)

`a)PTHH:`

`Fe + 2HCl -> FeCl_2 + H_2`

`0,15`  `0,3`             `0,15`     `0,15`       `(mol)`

`n_[Fe]=[8,4]/56=0,15(mol)`

`b)V_[H_2]=0,15.22,4=3,36(l)`

`c)V_[dd HCl]=[0,3]/[0,5]=0,6(l)`

`=>C_[M_[FeCl_2]]=[0,15]/[0,6]=0,25(M)`

\(a,n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

           0,15-->0,3----->0,15--->0,15

b, V_H2 = 0,15.22,4 = 3,36 (l)

\(c,V_{dd}=\dfrac{0,3}{0,5}=0,6\left(l\right)\\ \rightarrow C_{M\left(FeCl_2\right)}=\dfrac{0,15}{0,6}=0,25M\)

\(C\%ddKCl=\)\(\dfrac{20}{380+20}. 100 = 5\)\(\%\)

C%=20.100/380 = 5%

$C\%_{KCl}=\dfrac{20}{20+380}.100\%=5\%$

\(C\%ddKCl\)\(= \dfrac{20}{500} .100% = 4%\)\(\%\)=4\(\%\)

\(C\)\(\%\)\(=\dfrac{m_{ct}}{m_{dd}} .100\)\(\%\)= \(\dfrac{20}{500} . 100\)\(\%\)\(=4 \)\(\%\)

\(C\%=\dfrac{20}{500}.100=4\%\)

\(C\)\(\%=\)\( \dfrac{m_{ct}}{m_{dd}} . 100 \)\(\%\)\(= \dfrac{20}{500} . 100 \)\(\%\)\(=4\)\(\%\)

\(C\%_{NaCl}=\dfrac{15}{15+45}.100\%=25\%\)

\(∑m_{dd}= m_{NaCl} + m_{H_2O} = 15 + 45 = 60(gam)\)

\(\text{C%}ddNaCl=\)\(\dfrac{15}{60}.100\%=25\%\)