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\(n_{H_2SO_4}=\dfrac{19,6\%.500.1,12}{98}=1,12\left(mol\right)\\n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ MgO+H_2SO_4 \rightarrow MgSO_4+H_2O\\ CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2+H_2O\\ TH1:axit.hết\\ n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\\ \Rightarrow n_{MgO}=\dfrac{18-0,1.10}{40}=0,2\left(mol\right)\\ n_{H_2SO_4\left(p.ứ\right)}=0,2+0,1=0,3\left(mol\right)< 1,12\left(mol\right)\\ \Rightarrow LoạiTH1\\ TH2:axit.dư\\ \Rightarrow\left\{{}\begin{matrix}40a+100b=18\\b=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \Rightarrow\%m_{MgO}=\dfrac{0,2.40}{18}.100\approx44,444\%\Rightarrow\%m_{CaCO_3}\approx55,556\%\)
\(b,m_{ddB}=m_A+m_{ddH_2SO_4}-m_{CO_2}=18+500.1,12-0,1.44=573,6\left(g\right)\\ n_{H_2SO_4\left(dư\right)}=1,12-0,3=0,82\left(mol\right)\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{0,82.98}{573,6}.100\approx14,01\%\\ C\%_{ddCaCl_2}=\dfrac{0,1.111}{573,6}.100\approx1,935\%\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{573,6}.100\approx3,312\%\)
PTHH: \(CaO+2HNO_3\rightarrow Ca\left(NO_3\right)_2+H_2O\)
\(MgCO_3+2HNO_3\rightarrow Mg\left(NO_3\right)_2+CO_2\uparrow+H_2O\)
Ta có: \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{MgCO_3}=n_{Mg\left(NO_3\right)_2}\) \(\Rightarrow n_{CaO}=\dfrac{18-0,1\cdot84}{56}=\dfrac{6}{35}\left(mol\right)=n_{Ca\left(NO_3\right)_2}\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,1\cdot84}{18}\cdot100\%\approx46,67\%\\\%m_{CaO}=53,33\%\end{matrix}\right.\)
Theo đề bài, ta có: \(m_{ddHNO_3}=500\cdot1,08=540\left(g\right)\) \(\Rightarrow\Sigma n_{HNO_3}=\dfrac{540\cdot12,6\%}{63}=1,08\left(mol\right)\)
Theo PTHH: \(n_{HNO_3\left(p.ứ\right)}=2n_{CaO}+2n_{MgCO_3}=\dfrac{19}{35}\left(mol\right)\) \(\Rightarrow n_{HNO_3\left(dư\right)}=\dfrac{94}{175}\left(mol\right)\)
Mặt khác: \(m_{CO_2}=0,1\cdot44=4,4\left(g\right)\)
\(\Rightarrow m_{dd\left(sau.pư\right)}=m_A+m_{ddHNO_3}-m_{CO_2}=553,6\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Mg\left(NO_3\right)_2}=\dfrac{0,1\cdot148}{553,6}\cdot100\%\approx2,67\%\\C\%_{Ca\left(NO_3\right)_2}=\dfrac{\dfrac{6}{35}\cdot164}{553,6}\cdot100\%\approx5,08\%\\C\%_{HNO_3\left(dư\right)}=\dfrac{\dfrac{19}{35}\cdot63}{553,6}\cdot100\%\approx6,18\%\end{matrix}\right.\)
Thí nghiệm 1:
\(m_{ddH_2SO_4}=500\cdot1,12=560g\)
\(\Rightarrow m_{H_2SO_4}=\dfrac{560\cdot19,6\%}{100\%}=109,76g\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,1 0,3
Chất rắn không tan thu được là Ag.
Thí nghiệm 2:
\(n_{SO_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(BTe:3n_{Al}+n_{Ag}=2n_{SO_2}\)
\(\Rightarrow n_{Ag}=2\cdot0,4-3\cdot0,2=0,2mol\)
a)\(m_{Al}=0,2\cdot27=5,4g\)
\(m_{Ag}=0,2\cdot108=21,6g\)
b)Dung dịch B là \(Al_2\left(SO_4\right)_3\)
\(C_M=\dfrac{0,1}{0,5}=0,2M\)
a) \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
_____0,1<----------------0,1<------0,05
=> mNa = 0,1.23 = 2,3 (g)
=> \(\left\{{}\begin{matrix}\%Na=\dfrac{2,3}{4,7}.100\%=48,936\%\\\%Mg=100\%-48,936\%=51,064\%\end{matrix}\right.\)
b)
\(n_{Mg}=\dfrac{4,7-2,3}{24}=0,1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
______0,1-->0,2-------------->0,1
2Na + 2HCl --> 2NaCl + H2
0,1-->0,1-------------->0,05
=> mHCl = (0,1+0,2).36,5 = 10,95 (g)
=> \(C\%\left(HCl\right)=\dfrac{10,95}{200}.100\%=5,475\%\)
=> VH2 = (0,1 + 0,05).22,4 = 3,36 (l)