CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

a) \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

Theo PT: \(n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,1\times22,4=2,24\left(l\right)\)

\(m_{CO_2}=0,1\times44=4,4\left(g\right)\)

b) Theo PT: \(n_{HCl}=2n_{CaCO_3}=2\times0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,2\times36,5=7,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{7,3\times100\%}{14,6\%}=50\left(g\right)\)

c) Theo PT: \(n_{CaCl_2}=n_{CaCO_3}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaCl_2}=0,1\times111=11,1\left(g\right)\)

\(m_{dd}=10+50-4,4=55,6\left(g\right)\)

\(\Rightarrow C\%_{CaCl_2}=\dfrac{11,1}{55,6}\times100\%=19,96\%\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}84\cdot n_{MgCO_3}+100\cdot n_{CaCO_3}=18,4\\n_{MgCO_3}+n_{CaCO_3}=n_{CO_2}=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{MgCO_3}=0,1mol\\n_{CaCO_3}=0,1mol\end{matrix}\right.\)

\(\%m_{CaCO_3}=\dfrac{0,1\cdot100}{18,4}\cdot100\%=54,35\%\)

\(\%m_{MgCO_3}=100\%-54,35\%=45,65\%\)

a, \(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)

b, \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

Theo PT: \(n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\Rightarrow V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

c, \(n_{\left(CH_3COO\right)_2Ca}=n_{CaCO_3}=0,1\left(mol\right)\)

\(n_{CH_3COOH}=2n_{CaCO_3}=0,2\left(mol\right)\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{5\%}=240\left(g\right)\)

Ta có: m _{dd sau pư} = 10 + 240 - 0,1.44 = 245,6 (g)

\(\Rightarrow C\%_{\left(CH_3COO\right)_2Ca}=\dfrac{0,1.158}{245,6}.100\%\approx6,43\%\)

a)

$CaCO_3 + 2CH_3COOH \to (CH_3COO)_2Ca + CO_2 + H_2O$
b)

$n_{CaCO_3} = \dfrac{10}{100} = 0,1(mol)$
Theo PTHH : 

$n_{CO_2} = n_{CaCO_3} = 0,1(mol)$
$V_{CO_2} = 0,1.22,4 = 2,24(lít)$

c)

$n_{CH_3COOH} = 2n_{CaCO_3} = 0,2(mol)$
$m_{dd\ CH_3COOH} = \dfrac{0,2.60}{5\%} = 240(gam)$
$m_{dd\ sau\ pư} = 10 + 240 - 0,1.44 = 245,6(gam)$
$n_{(CH_3COO)_2Ca} = n_{CaCO_3} = 0,1(mol)$
$C\%_{(CH_3COO)_2Ca} = \dfrac{0,1.158}{245,6}.100\% = 6,4\%$

\(n_{Zn}=\dfrac{65}{65}=1mol\)

\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

 1    <   0,8                              ( mol )

0,4        0,8           0,4        0,4    ( mol )

\(m_{ddspứ}=200+65-0,4.2=264,2g\)

\(\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,4.136}{264,2}.100=14,93\%\\C\%_{H_2}=\dfrac{0,4.2}{264,2}.100=0,3\%\\C\%_{Zn\left(dư\right)}=\dfrac{\left(1-0,4\right).65}{264,2}.100=14,76\%\end{matrix}\right.\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)

\(m_{HCl}=\dfrac{150\cdot14,6\%}{100\%}=21,9g\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6mol\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,2       0,6           0           0

0,2       0,4           0,2        0,2

0          0,2           0,2        0,2

Sau phản ứng, \(ddHCl\) còn dư và dư 0,2mol.

\(V_{H_2}=0,2\cdot22,4=4,48l\)

\(a)n_{CaCO_3}=\dfrac{10}{100}=0,1mol\\ n_{H_2SO_4}=\dfrac{150.19,6\%}{100\%.98}=0,3mol\\ CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2+H_2O\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,3}{1}\Rightarrow H_2SO_4.dư\\ n_{CO_2}=n_{CaSO_4}=n_{H_2SO_4.pứ}=n_{CaCO_3}=0,1mol\\ V_{CO_2}=0,1.22,4=2,24l\\ b)m_{dd.sau.pứ}=10+150-0,1.44=155,6g\\ C_{\%CaSO_4}=\dfrac{0,1.136}{155,6}\cdot100\%=8,74\%\\ C_{\%H_2SO_4.dư}=\dfrac{\left(0,3-0,1\right).98}{155,6}\cdot100\%=12,6\%\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(m_{HCl}=21,9g\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\)

=> HCl dư

\(\Rightarrow n_{H_2}=0,2mol\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48l\)

bổ sung ý b)

Khối lượng dung dịch sau phản ứng = mZn + mHCl - mH₂ thoát ra = 13 +150 - 0,2 .2 = 162,6 gam

Dung dịch thu được sau phản ứng gồm \(\left\{{}\begin{matrix}ZnCl_2\\HCl_{dư}\end{matrix}\right.\)

nZnCl₂ = nZn = 0,2 mol => mZnCl₂ = 0,2 . 136 = 27,2 gam

=> C% ZnCl₂ = \(\dfrac{27,2}{162,6}\).100= 16,72%

nHCl dư = 0,6 - 0,4 = 0,2 mol 

mHCl dư= 0,2.36,5 = 7,3 gam

=> C% HCl dư = \(\dfrac{7,3}{162,6}\).100 = 4,5%

CaCO₃ + 2HCl -> CaCl₂ + CO₂ + H₂O

n_CaCO3=0,1(mol)

Theo PTHH ta có:

n_HCl=2n_CaCO3=0,2(mol)

n_CaCl2=n_CaCO3=0,1(mol)

n_CO2=n_CaCO3=0,1(mol)

m_HCl=36,5.0,2=7,3(g)

m _{dd HCl}=7,3:20%=36,5(g)

V_{dd HCl}=36,5.:1,1=33,2(ml)

m_dd =10+36,5-44.0,1=42,1(g)

m_CaCl2=111.0,1=11,1(g)

C% dd CaCl₂=\(\dfrac{11,1}{42,1}.100\%=26,36\%\)

a)CaCO3 +2HCl -->CaCl2 +CO2 +H2O (1)

b) nCaCO3=0,1(mol)

theo (1): nHCl=nCaco3=0,2(mol)

mddHCl=36,5(g)

=>VHCl=36,5/1,1=33,18(l)

c) theo (1):nCO2=nCaCl2=nCaCO3=0,1(mol)

=>mCaCl2=11,1(g)

mdd sau pư=10+36,5-0,1.44=42,1(g)

=.C%ddCaCl2=26,3(%)

Ta có: \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

____0,4_____0,8____0,4_____0,4 (mol)

a, \(m_{HCl}=0,8.36,5=29,2\left(g\right)\)

b, \(m_{FeCl_2}=0,4.127=50,8\left(g\right)\)

c, \(m_{H_2}=0,4.2=0,8\left(g\right)\)

d, \(m_{ddHCl}=\dfrac{29,2}{14,6\%}=200\left(g\right)\)

m _{dd sau pư} = 22,4 + 200 - 0,8 = 221,6 (g)

e, \(C\%_{FeCl_2}=\dfrac{50,8}{221,6}.100\%\approx22,92\%\)