Bài 42: Nồng độ dung dịch

\(n_{Na}=\dfrac{1,15}{23}=0,05\left(mol\right)\)

PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

          0,05------------>0,05------>0,025

\(\rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,05.40+120.15\%=20\left(g\right)\\m_{dd}=1,15+120-0,025.2=121,1\left(g\right)\end{matrix}\right.\\ \rightarrow C\%_{NaOH}=\dfrac{20}{121,1}.100\%=16,12\%\)

`n_[CuO]=[0,8]/80=0,01(mol)`

`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`

`0,01`   `0,01`        `0,01`                    `(mol)`

`a)m_[Cu]=0,01.64=0,64(g)`

`b)V_[H_2]=0,01.22,4=0,224(l)`

`c)`

`Fe + 2HCl -> FeCl_2 + H_2`

`0,01`  `0,02`                       `0,01`       `(mol)`

`@m_[Fe]=0,01.56=0,56(g)`

`@m_[dd HCl]=[0,02.36,5]/20 . 100=3,65(g)`

\(C\%_{KNO_3}=\dfrac{31,6}{100+31,6}.100\%=24,01\%\)

\(1,n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

           0,3---->0,6------------------>0,3

\(2,C_{M\left(HCl\right)}=\dfrac{0,6}{0,3}=2M\\ 3,V_{H_2}=0,3.22,4=6,72\left(l\right)\)

\(C_{M\left(KCl\right)}=\dfrac{0,1}{0,5}=0,2M\)

Cm(kcl)=\(\dfrac{0,1}{0,5}\)=0,2 M

\(C\%_{NaCl}=\dfrac{10}{10+40}.100\%=20\%\)

Viết theo lần lượt nhé: \(FeO,Fe_2O_3,BaO,K_2O\)

`a)n_[NaCl]=0,75.0,5=0,375(mol)`

`=>m_[NaCl]=0,375.58,5=21,9375(g)`

`b)n_[KNO_3]=2.0,5=1(mol)`

`=>m_[KNO_3]=1.101=101(g)`

`c)n_[CaCl_2]=0,1.0,25=0,025(mol)`

`=>m_[CaCl_2]=0,025.111=2,775(g)`

`d)n_[Na_2 SO_4]=0,3.1,5=0,45(mol)`

`=>m_[Na_2 SO_4]=0,45.142=63,9(g)`

\(V_{ddKCl}=\dfrac{0,15}{2}=0,075\left(l\right)\)

\(m_{dd}=\dfrac{40}{20\%}=200\left(g\right)\)