\(n_{SO_3}=\dfrac{16}{80}=0,2\left(mol\right)\)

\(\Rightarrow n_O=0,2.3=0,6\left(mol\right)\)

\(n_{O_2}=0,6:2=0,3\left(mol\right)\)

Khí là sunfuro còn axit mới là axit sunfuric chứ

a. Chất rắn từ đen chuyển dần sang đỏ gạch, đồng thời hơi nước bám lên thành ống thuỷ tinh.

\(b.n_{CuO\left(pư\right)}=a\left(mol\right)=n_{\left[O\right]pư}\\ CuO+H_2-^{t^0}->Cu+H_2O\\ m_{rắn}=20-16a=16,8\\ a=0,2\left(mol\right)=n_{H_2}\\ V_{H_2}=0,2.22,4=4,48\left(L\right)\)

Tổng quát: 

\(OH^{^-}+H^{^+}->H_2O\\ Ba^{^{2+}}+SO_4^{^{2-}}->BaSO_4\downarrow\\ n_{OH^-}=2.1,2.0,2=0,48mol=n_{H^+}\\ m_{ddX}=a\left(g\right)\\ n_{H^+}=0,48=\dfrac{0,146a}{36,5}+\dfrac{0,196a}{98}.2\\ a=60\left(g\right)\\ m_{BaSO_4}=233.1,2.0,2=55,92g\)

\(n_{BaCl_2}=\dfrac{4,16}{208}=0,02 \left(mol\right)\)

\(n_{AgNO_3}=\dfrac{3,4}{170}=0,02\left(mol\right)\)

\(BaCl_2+2AgNO_3\rightarrow2AgCl+Ba\left(NO_3\right)_2\)

0,01 <----- 0,02 --------------------> 0,01

Do \(\dfrac{0,02}{1}>\dfrac{0,02}{2}\) => \(BaCl_2\) dư và dư: \(0,02-0,01=0,01\left(mol\right)\)

b.

Dung dịch sau phản ứng gồm \(BaCl_2\) dư và \(Ba\left(NO_3\right)_2\)

c.

\(CM_{Ba\left(NO_3\right)_2}=\dfrac{0,01}{0,2}=0,05M\)

\(CM_{BaCl_2}=\dfrac{0,01}{0,2}=0,05M\)

\(V_{Cl_2,đktc}=0,05\cdot22,4=1,12\left(L\right)\)

a, \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

b, Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 160y = 32 (1)

Theo PT: \(\left\{{}\begin{matrix}n_{CuCl_2}=n_{Cu}=x\left(mol\right)\\n_{FeCl_3}=2n_{Fe_2O_3}=2y\left(mol\right)\end{matrix}\right.\) ⇒ 135x + 325y = 59,5 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2.80=16\left(g\right)\\m_{Fe_2O_3}=0,1.160=16\left(g\right)\end{matrix}\right.\)

c, Theo PT: \(n_{HCl}=2n_{CuO}+6n_{Fe_2O_3}=1\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{1}{0,5}=2\left(l\right)\)

a, \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,2.84}{64,8}.100\%\approx25,93\%\\\%m_{MgSO_4}\approx74,07\%\end{matrix}\right.\)

b, - Dung dịch C gồm: MgCl₂, MgSO₄ và HCl dư.

Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{CO_2}=0,2\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{CO_2}=0,4\left(mol\right)\end{matrix}\right.\)

Ta có: \(m_{HCl}=100.18,25\%=18,25\left(g\right)\Rightarrow n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\)

\(m_{MgSO_4}=64,8-0,2.84=48\left(g\right)\Rightarrow n_{MgSO_4}=\dfrac{48}{120}=0,4\left(mol\right)\)

Có: m _{dd sau pư} = 64,8 + 100 - 0,2.44 = 156 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,2.95}{156}.100\%\approx12,18\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{156}.100\%\approx2,34\%\\C\%_{MgSO_4}=\dfrac{48}{156}.100\%\approx30,77\%\end{matrix}\right.\)

c, PT: \(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)

\(HCl+NaOH\rightarrow NaCl+H_2O\)

\(MgSO_4+2NaOH\rightarrow Na_2SO_4+Mg\left(OH\right)_{2\downarrow}\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

Theo PT: \(n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}+n_{MgSO_4}=0,6\left(mol\right)\)

\(\Rightarrow m_{cr}=m_{MgO}=0,6.40=24\left(g\right)\)

Ta có: \(n_{HCl}=0,5.0,5=0,25\left(mol\right)\) \(\Rightarrow n_{HCl\left(pư\right)}=0,25.80\%=0,2\left(mol\right)\)

Gọi KL cần tìm là A. 

\(n_A=\dfrac{1}{M_A}\left(mol\right)\)

PT: \(A+2HCl\rightarrow ACl_2+H_2\)

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

Mà: Sự hòa tan vẫn còn tiếp diễn.

\(\Rightarrow\dfrac{1}{M_A}>0,1\Rightarrow M_A< 10\left(g/mol\right)\)

Vậy: A là Beri. (Be)

\(n_{Na_2CO_3}=\dfrac{200.0,159}{106}=0,3mol\\ n_{BaCl_2}=\dfrac{200.0,208}{208}=0,2mol\\ a.Na_2CO_3+BaCl_2->2NaCl+BaCO_3\\ n_{Na_2CO_3}:1>n_{BaCl_2}:1\\ m_B=197.0,2=39,4g\\ Na_2CO_3+2HCl->2NaCl+H_2O+CO_2\\ V=\dfrac{2.0,1}{1}=0,2\left(L\right)=200\left(mL\right)\\ b.m_A=200+200-39,4=360,6g\\ C\%_{Na_2CO_3du}=\dfrac{106.0,1}{360,6}.100\%=2,94\%\\ C\%_{NaCl}=\dfrac{58,5.0,4}{360,6}.100\%=6,49\%\)

a) 

\(Na_2CO_3+BaCl_2\rightarrow BaCO_3+2NaCl\)

0,2 <---------- 0,2 ------> 0,2 -----> 0,4

\(n_{Na_2CO_3}=\dfrac{200.15,9\%}{100\%}:106=0,3\left(mol\right)\)

\(n_{BaCl_2}=\dfrac{200.20,8\%}{100\%}:208=0,2\left(mol\right)\)

Do \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\) nên \(Na_2CO_3\) dư sau phản ứng.

Dung dịch A: \(n_{Na_2CO_3}=0,3-0,2=0,1\left(mol\right);n_{NaCl}:0,4\left(mol\right)\)

Kết tủa B: \(BaCO_3\)

\(m_B=m_{BaCO_3}=0,2.197=39,4\left(g\right)\)

Dung dịch A td với HCl:

\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)

0,1 ---------> 0,2

\(V=V_{HCl}=\dfrac{0,2}{1}=0,2\left(l\right)\)

b)

\(m_{dd}=m_{dd.Na_2CO_3}+m_{dd.BaCl_2}-m_{BaCO_3}=200+200-39,4=360,6\left(g\right)\)

\(C\%_{Na_2CO_3}=\dfrac{0,1.106.100\%}{360,6}=2,94\%\)

\(C\%_{NaCl}=\dfrac{0,4.58,5.100\%}{360,6}=6,49\%\)