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\(a)n_{Ba\left(OH\right)_2}=0,05\cdot0,2\cdot2=0,02mol\\ pH=1\Rightarrow\left[OH^-\right]=0,1M\Rightarrow n_{HCl}=0,1\cdot0,3=0,03mol\\ n_{Ba\left(OH\right)_2}+n_{HCl}=0,02+0,03=0,05mol\\ \Rightarrow C_M=\dfrac{0,05}{0,5}=0,1M\Rightarrow pH=1\)
\(n_{C\left(1\right)}=n_{CaCO_3\left(1\right)}=\dfrac{20}{100}=0,02\left(mol\right)\)
\(n_{C\left(2\right)}=n_{HCO_3^-}=2n_{CaCO_3\left(2\right)}=2.\dfrac{10}{100}=0,2\left(mol\right)\)
\(\Rightarrow\Sigma n_C=0,22\left(mol\right)\)
\(\Rightarrow V_{CO_2}=4,928\left(l\right)\)
C
2Fe(NO3)3 + 3Na2CO3 + 3H2O --> 2Fe(OH)3\(\downarrow\) + 6NaNO3 + 3CO2
2Fe(OH)3 --to--> Fe2O3 + 3H2O
Ban đầu:\(n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,25.\left(0,08+2.0,01\right)=0,025\left(mol\right)\)
\(n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,25a.2=0,5a\)
\(n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,25a\); \(n_{SO_4^{2-}}=n_{H_2SO_4}=0,0025\left(mol\right)\)
Dung dịch sau phản ứng có pH = 12 => pOH = 2
\(\Rightarrow\left[OH^-\right]=10^{-2}=0,01\left(M\right)\Rightarrow n_{OH^-}=0,01.0,5=0,005\left(mol\right)\)Vì pH = 12 > 7 nên \(H^+\) hết, \(OH^-\) còn.
\(H^++OH^-\rightarrow H_2O\)
0,025-->0,025
=> \(n_{OH^-}\text{còn}=0,5a-0,025=0,005\Rightarrow a=0,06\left(mol\text{/}l\right)\)
Từ đó suy ra được \(n_{Ba^{2+}}=0,25.0,06=0,015\left(mol\right)\)
\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)
0,0025<--0,0025
=> \(m=m_{BaSO_4}=0,0025.233=0,5825\left(gam\right)\)
\(n_{CaCO_3}=\dfrac{10}{100}=0.1\left(mol\right)\)
\(TH1:\)
\(n_{CO_2}=n_{CaCO_3}=0.1\left(mol\right)\)
\(V=0.1\cdot22.4=2.24\left(l\right)\)
\(TH2:\)
\(n_{CO_2}=n_{CaCO_3}+2n_{Ca\left(HCO_3\right)_2}=0.1+2\cdot\left(0.15-0.1\right)=0.2\left(mol\right)\)
\(V=0.2\cdot22.4=4.48\left(l\right)\)
\(\Rightarrow2.24\left(l\right)< V< 4.48\left(l\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{OH^-}=0,1+0,1\cdot2=0,3\left(mol\right)\\n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối
PT: \(CO_2+OH^-\rightarrow HCO_3^-\)
a_____a_________a (mol)
\(CO_2+2OH^-\rightarrow CO_3^{2-}+H_2O\)
b____2b________b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}a+b=0,2\\a+2b=0,3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
Mặt khác: \(n_{Ba^{2+}}=0,1\left(mol\right)=n_{CO_3^{2-}}=n_{BaCO_3}\) \(\Rightarrow m_{BaCO_3}=0,1\cdot197=19,7\left(g\right)\)
C
\(Ca\left(OH\right)_2+Ca\left(HCO_3\right)_2->2CaCO_3\downarrow+2H_2O\)