Bài 2 : 

\(n_{CaCO3}=\dfrac{20}{100}=0,2\left(mol\right)\)

\(m_{ct}=\dfrac{20.91,25}{100}=18,25\left(g\right)\)

\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

Pt : \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O|\)

           1              2              1           1          1

         0,2           0,5           0,2        0,2

Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\)

                     ⇒ CaCO₃ phản ứng hết , HCl dư

                    ⇒ Tính toán dựa vào số mol của CaCO₃

\(n_{CO2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{CO2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

\(n_{CaCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CaCl2}=0,2.111=22,2\left(g\right)\)

\(n_{HCl\left(dư\right)}=0,5-\left(0,2.2\right)=0,1\left(mol\right)\)

⇒ \(m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

\(m_{ddspu}=20+91,25-\left(0,2.44\right)=102,45\left(g\right)\)

\(C_{CaCl2}=\dfrac{22,2.100}{102,45}=21,67\)⁰/₀

\(C_{HCl\left(dư\right)}=\dfrac{3,65.100}{102,45}=3,56\)⁰/₀

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Ta có: \(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\)

Ta lại có: \(C_{\%_{HCl}}=\dfrac{m_{HCl}}{91,25}.100\%=20\%\)

=> m_HCl = 18,25 (g)

=> \(n_{HCl}=\dfrac{18,25}{36,5}=0,2\left(mol\right)\)

PTHH: CaCO₃ + 2HCl ---> CaCl₂ + H₂O + CO₂↑

Ta thấy: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\)

Vậy CaCO₃ dư

Theo PT: \(n_{CO_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

=> \(V_{CO_2}=0,1.22,4=2,24\left(lít\right)\)

Theo PT: \(n_{H_2O}=n_{CaCl_2}=n_{H_2O}=0,1\left(mol\right)\)

=> \(m_{CaCl_2}=0,1.111=11,1\left(g\right)\)

\(m_{H_2O}=0,1.18=1,8\left(g\right)\)

=> \(m_{dd_{CaCl_2}}=20+91,25-1,8-\left(44.0,1\right)=105,05\left(g\right)\)

=> \(C_{\%_{CaCl_2}}=\dfrac{11,1}{105,05}.100\%=10,57\%\)

Ta có: \(m_{dd_{H_2O}}=20+91,25-11,1-\left(44.0,1\right)=95,75\left(g\right)\)

=> \(C_{\%_{H_2O}}=\dfrac{1,8}{95,75}.100\%=1,88\%\)

Ta có: \(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)

\(PTHH:2Na+2H_2O--->2NaOH+H_2\uparrow\)

a. Theo PT: \(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(lít\right)\)

b. Theo PT: \(n_{NaOH}=n_{Na}=0,4\left(mol\right)\)

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

Ta có: \(m_{dd_{NaOH}}=9,2+200-0,2.2=208,8\left(g\right)\)

\(\Rightarrow C_{\%_{NaOH}}=\dfrac{16}{208,8}.100\%=7,66\%\)

pika pikacha pikachu pi pikapikcha

     pika+pikachu=(pikapika+pipika)

                                                                    Ònc lịa Pikachu hkôgn btếi

a) \(n_{FeCl_3}=\dfrac{16,25}{162,5}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

            0,05<----0,3<-----0,1

=> \(m_{Fe_2O_3}=0,05.160=8\left(g\right)\)

b) 

\(m_{HCl\left(bd\right)}=91,25.16\%=14,6\left(g\right)\)

m_{dd sau pư} = 8 + 91,25 = 99,25 (g)

\(\left\{{}\begin{matrix}C\%\left(FeCl_3\right)=\dfrac{16,25}{99,25}.100\%=16,373\%\\C\%\left(HCldư\right)=\dfrac{14,6-0,3.36,5}{99,25}.100\%=3,678\%\end{matrix}\right.\)

\(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\\ n_{H_2O}=\dfrac{15}{18}=\dfrac{5}{6}\left(mol\right)\)

PTHH: 2Na + 2H₂O ---> 2NaOH + H₂

LTL: \(2>\dfrac{5}{6}\) => Na dư

Theo pthh: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{1}{2}n_{H_2O}=\dfrac{1}{2}.\dfrac{5}{6}=\dfrac{5}{12}\left(mol\right)\\n_{Na\left(pư\right)}=n_{NaOH}=n_{H_2O}=\dfrac{5}{6}\left(mol\right)\end{matrix}\right.\)

=> \(V_{H_2}=\dfrac{5}{12}.22,4=\dfrac{28}{3}\left(l\right)\)

\(m_{dd}=15+23.\dfrac{5}{6}-\dfrac{5}{12}.2=\dfrac{100}{3}\\ m_{NaOH}=\dfrac{5}{6}.40=\dfrac{100}{3}\left(g\right)\\ \rightarrow C\%_{NaOH}=\dfrac{\dfrac{100}{3}}{\dfrac{100}{3}}.100\%=100\%\)

\(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\\ n_{H_2O}=\dfrac{15}{18}=0,83\left(mol\right)\\ pthh:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\) 
                    0,83             0,83             0,416 
\(V_{H_2}=0,416.22,4=9,3l\\ m_{\text{dd}}=46+15-\left(0,416.2\right)=60,17\left(g\right)C\%=\dfrac{0,83.40}{60,17}.100\%=55,176 \%\)