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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{MgCO_3}=\dfrac{16,8}{84}=0,2mol\\ a.MgCO_3+H_2SO_4->MgSO_4+H_2O+CO_2\\ 2NaOH+H_2SO_{\text{4 }}->Na_2SO_4+2H_2O\\ b.n_{H_2SO_4dư}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.80.0,1:40=0,1mol\\ n_{H_2SO_4\left(MgCO_3\right)}=0,2mol\\ c.C\%=\dfrac{98.0,3}{200}.100\%=14,7\%\\ V=0,2.22,4=4,48L\\ d.m_{ddsau}=200+16,8-44.0,2+80=288g\\ C\%_{Na_2SO_4}=\dfrac{40.0,1}{288}.100\%=1,39\%\\ C\%_{MgSO_4}=\dfrac{120.0,2}{288}.100\%=8,33\%\)
\(n_{MgCO_3}=\dfrac{16,8}{84}=0,2\left(mol\right)\)
\(n_{NaOH}=\dfrac{80}{40}=2\left(mol\right)\)
PTHH :
\(MgCO_3+H_2SO_4\rightarrow MgSO_4+H_2O+CO_2\uparrow\)
0,2 0,2 0,2
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
2 1 1
Vậy có 0,2 mol H2SO4 phản ứng với MgCO3
có 1 mol H2SO4 phản ứng với NaOH
\(m_{H_2SO_4}=1,2.98=117,6\left(g\right)\)
\(c,C\%_{H_2SO_4}=\dfrac{117,6}{200}.100\%=58,8\%\)
\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
\(d,m_{Na_2SO_4}=1.142=142\left(g\right)\)
\(m_{ddNaOH}=\dfrac{80.100}{10}=800\left(g\right)\)
\(m_{ddH_2SO_4dư}=1.98:58,8\%\approx166,67\left(g\right)\)
\(m_{ddNa_2SO_4}=800+166,67=966,67\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{142}{966,67}.100\%\approx14,69\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo PT: \(n_{FeSO_4}=n_{H_2}=n_{Fe}=0,01\left(mol\right)\)
\(\Rightarrow m_{FeSO_4}=0,01.152=1,52\left(g\right)\)
\(V_{H_2}=0,01.22,4=0,224\left(l\right)\)
b, \(n_{H_2SO_4}=n_{Fe}=0,01\left(mol\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{0,01.98}{19,6\%}=5\left(g\right)\)
c, Ta có: m dd sau pư = 0,56 + 5 - 0,01.2 = 5,54 (g)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{1,52}{5,54}.100\%\approx27,44\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Na2O + H2O----> 2NaOH
nNa2O = 6,2/(23.2 + 16)=0,1
=> nNaOH = 0,1.2=0,2
CM dd NaOH =0,2/0,1=2M
b, mdd=1,05.100=105 g
=> C% dd= 0,2.40.100/105=7,61%
\(n_{NaCl}=\dfrac{35,1}{58,5}=0,6\left(mol\right)\)
Pt: \(2NaCl+H_2SO_4\rightarrow Na_2SO_4+2HCl\)
\(\Rightarrow n_{HCl}=n_{NaCl}=0,6\left(mol\right)\)
\(\Sigma_{m_{dd\left(spu\right)}}=m_{HCl}+m_{H_2O}=0,6.36,5+78,1.1=100\left(g\right)\)
\(C\%_{HCl}=\dfrac{0,6.36,5}{100}.100=21,9\%\)
\(V_{HCl}=\dfrac{100}{12}=\dfrac{25}{3}\left(l\right)\)
\(C_{M_{HCl}}=\dfrac{0,6}{\dfrac{25}{3}}=0,072M\)
Câu b giải cũng tương tự thôi, tính m ra -> số mol..