\(a,PTHH:KHCO_3+2H_2SO_4\rightarrow K_2SO_4+2CO_2\uparrow+2H_2O\\ b,n_{KHCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\\ \Rightarrow n_{CO_2}=2n_{KHCO_3}=0,4\left(mol\right)\\ \Rightarrow V_{CO_2}=0,4\cdot22,4=8,9\left(l\right)\\ c,n_{H_2SO_4}=n_{CO_2}=0,4\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,4\cdot98=39,2\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{39,2\cdot100\%}{19,6\%}=200\left(g\right)\)

a) PTHH: Zn + H2SO4 -> ZnSO4 + H2

nH2= 0,15(mol)

=> nZn=nH2SO4=nZnSO4=nH2=0,15(mol)

b) mZn=0,15.65=9,75(g)

c) CMddH2SO4= 0,15/ 0,05=3(M)

d) mZnSO4= 161. 0,15=24,15(g)

200ml = 0,2l

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

        1         2              1           1

     0,3      0,6             0,3        0,3

a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)

              1            1              1            1

            0,6          0,6

\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)

\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)

 Chúc bạn học tốt

a) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

b+c)

Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)=n_{H_2SO_4}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{ddH_2SO_4}=\dfrac{0,3\cdot98}{30\%}=98\left(g\right)\end{matrix}\right.\)

d) PTHH: \(ZnSO_4+BaCl_2\rightarrow ZnCl_2+BaSO_4\downarrow\)

Ta có: \(n_{BaCl_2}=\dfrac{260\cdot20\%}{208}=0,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,25}{1}\) \(\Rightarrow\) ZnSO₄ còn dư, BaCl₂ phản ứng hết

\(\Rightarrow\left\{{}\begin{matrix}n_{ZnCl_2}=0,25mol=n_{BaSO_4}\\n_{ZnSO_4\left(dư\right)}=0,05mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,25\cdot136=34\left(g\right)\\m_{BaSO_4}=0,25\cdot233=58,25\left(g\right)\\m_{ZnSO_4\left(dư\right)}=0,05\cdot161=8,05\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{H_2}=0,3\cdot2=0,6\left(g\right)\)

\(\Rightarrow m_{dd}=m_{Zn}+m_{ddH_2SO_4}-m_{H_2}+m_{ddBaCl_2}-m_{BaSO_4}=318,65\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{34}{318,65}\cdot100\%\approx10,67\%\\C\%_{ZnSO_4\left(dư\right)}=\dfrac{8,05}{318,65}\cdot100\%\approx2,53\%\end{matrix}\right.\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

(mol)_____0,2____0,2______0,2____0,2__

\(a.V_{H_2}=22,4.0,2=4,48\left(l\right)\)

\(b.m_{ddH_2SO_4}=\dfrac{0,2.98.100}{24,5}=80\left(g\right)\)

\(c.m_{ddspu}=13+80-0,2.2=92,6\left(g\right)\\ \Rightarrow C\%_{ddspu}=\dfrac{0,2.136}{92,6}.100=29,4\left(\%\right)\)

PTHH: \(Mg+H_2SO_4\rightarrow MgsO_4+H_2\uparrow\)

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)=n_{MgSO_4}=n_{H_2SO_4}\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgSO_4}=0,2\cdot120=24\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\\C\%_{H_2SO_4}=\dfrac{0,2\cdot98}{294}\cdot100\%\approx6,67\%\end{matrix}\right.\)