a, PT: \(Na_2SO_3+2HCl\rightarrow2NaCl+SO_2+H_2O\)

Ta có: \(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)\)

Theo PT: \(n_{SO_2}=n_{Na_2SO_3}=0,1\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(n_{NaCl}=n_{HCl}=2n_{Na_2SO_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{NaCl}=0,2.58,5=11,7\left(g\right)\)

c, \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{7,3}{10\%}=73\left(g\right)\)

d, Ta có: m _{dd sau pư} = 12,6 + 73 - 0,1.64 = 79,2 (g)

\(\Rightarrow C\%_{NaCl}=\dfrac{11,7}{79,2}.100\%\approx14,77\%\)

a) $Fe + H_2SO_4 \to FeSO_4 + H_2$

b) 

Theo PTHH : $n_{H_2SO_4} = n_{H_2} = \dfrac{224}{1000.22,4} = 0,01(mol)$
$C\%_{H_2SO_4} = \dfrac{0,01.98}{100}.100\% = 0,98\%$

$\Rightarrow x = 0,98$

c) $n_{Fe} = n_{H_2} = 0,01(mol)$
$m_{dd\ sau\ pư} = m_{Fe} + m_{dd\ H_2SO_4} - m_{H_2} = 0,01.56 + 100 - 0,01.2 = 100,54(gam)$
$C\%_{FeSO_4} = \dfrac{0,01.152}{100,54}.100\% = 1,51\%$

\(a.CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ n_{CaCO_3}=n_{CO_2}=\dfrac{448:1000}{22,4}=0,02\left(mol\right)\\ n_{HCl}=0,02.2=0,04\left(mol\right)\\ C\%_{ddHCl}=\dfrac{0,04.36,5}{1,18.200}\approx0,619\%\\b.m_{CaCO_3}=0,02.100=2\left(g\right)\\ \%m_{CaCO_3}=\dfrac{2}{5}.100=40\%\\ \%m_{CaSO_4}=100\%-40\%=60\% \)

Mình tra KLR của dd HCl trên mạng là 1,18g/ml nên áp dụng vào bài nha ^^

ta có 200cm3=0,2 lítn hcl=2*0,2=0,4 molgọi số mol của caco3 là a,na2co3 là bcaco3 + 2hcl -> cacl2 + co2 + h2oa(mol)---2a(mol)--a-------a--------ana2co3 + 2hcl -> 2nacl + co2 + h2ob(mol)---2b(mol)---2b-------b------bta có100a+106b=20,62a+2b=0,4=> a=b=0,1 mol=> m caco3=10g; m na2co3=10,6 g 

PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)  (1)

            \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)  (2)

a) Ta có: \(\Sigma n_{HCl}=0,2\cdot2=0,4\left(mol\right)\)

Gọi số mol của Na₂CO₃ là \(a\) \(\Rightarrow n_{HCl\left(1\right)}=2a\left(mol\right)\)

Gọi số mol của CaCO₃ là \(b\) \(\Rightarrow n_{HCl\left(2\right)}=2b\left(mol\right)\)

Ta lập được hệ phương trình:

\(\left\{{}\begin{matrix}2a+2b=0,4\\106a+100b=20,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,1\cdot100=10\left(g\right)\\m_{Na_2CO_3}=10,6\left(g\right)\end{matrix}\right.\)

b) Theo PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}=0,2mol\\n_{CaCl_2}=n_{CaCO_3}=0,1mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\end{matrix}\right.\)

Mặt khác: \(\left\{{}\begin{matrix}m_{CO_2}=0,2\cdot44=8,8\left(g\right)\\m_{ddHCl}=200\cdot1,2=240\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd\left(saup/ư\right)}=m_{hh}+m_{ddHCl}-m_{CO_2}=251,8\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{251,8}\cdot100\%\approx4,65\%\\C\%_{CaCl_2}=\dfrac{11,1}{251,8}\cdot100\%\approx4,41\%\end{matrix}\right.\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1           0,2            0,1      0,1 
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\ V_{H_2}=0,1.22,4=2,24l\\ m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)  
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2             0,4       0,2              0,2 
\(m_{HCl}=0,4.36,5=14,6g\\ V_{H_2}=0,2.22,4=4,48l\\ m\text{dd}=4,8+200-0,4=204,4g\\ C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)

1,2 lít chắc ở ĐK thường ha, cứ cho vậy đi còn không 1,12l ở đktc ha.

nCO2= 0,05(mol)

PTHH:   2 NaOH + CO2 -> Na2CO3 + H2O

nNaOH= 0,1(mol); nNa2CO3= 0,05(mol)

mNaOH= 4(g); mNa2CO3=5,3(g)

mddNaOH= 4: 6%= 200/3(g)

=>mddNa2CO3= mCO2+ mddNaOH = 0,05.44+ 200/3= 1033/15(g)

=> C%ddNa2CO3= [5,3/(1033/15)].100=7,696%

a) $Zn+ 2HCl \to ZnCl_2 + H_2$
b) $n_{HCl} = \dfrac{250.7,3\%}{36,5} = 0,5(mol)$
$n_{Zn} = n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,25(mol)$
$m = 0,25.65 =16,25(gam) ; V_{H_2} = 0,25.22,4 = 5,6(lít)$
c)

$m_{dd\ sau\ pư} = 16,25 + 250 - 0,25.2 = 265,75(gam)$
$C\%_{ZnCl_2}  = \dfrac{0,25.136}{265,75}.100\% = 12,8\%$

\(a/ 4Zn+2HCl \to ZnCl_2+H_2 \\ n_{HCl}=\frac{250.7,3\%}{36,5}=0,5(mol)\\ b/ \\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=\frac{1}{2}.n_{HCl}=\frac{1}{2}.0,5=0,25(mol)\\ m_{Zn}=0,25.65=16,25(g)\\ V_{H_2}=0,25.22,4=5,6(l)\\ c/ \\ C\%_{ZnCl_2}=\frac{0,25.136}{16,25+250-0,25.2}.100=12,8\% \)

a)

 
b) 

nZn=nH2=12nHCl=0,25(mol)
m=0,25.65=16,25(gam);VH2=0,25.22,4=5,6(lít)
c)

mdd sau pư=16,25+250−0,25.2=265,75(gam)