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

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

0,2       0,4         0,2        0,2

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

\(m_{ZnCl_2}=0,2\cdot136=27,2\left(g\right)\)

\(m_{H_2}=0,2\cdot2=0,4\left(g\right)\)

\(m_{ddsau}=13+200-0,4=212,6\left(g\right)\)

\(\Rightarrow C\%=\dfrac{27,2}{212,6}\cdot100\%=12,8\%\)

Zn+2HCl->ZnCl2+H2

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

n Zn=\(\dfrac{19,5}{65}=0,3mol\)

=>VH2=0,3.22,4=6,72l

=>m HCl=0,6.36,5=21,9g

a)

Gọi $n_{NaOH} = a(mol) ; n_{KOH} = b(mol) \Rightarrow 40a + 56b = 3,04(1)$
$NaOH + HCl \to NaCl + H_2O$
$KOH + HCl \to KCl + H_2O$
$m_{muối} = 58,5a + 74,5b = 4,15(2)$

Từ (1)(2) suy ra a = 0,02 ; b = 0,04

$n_{HCl} = a + b = 0,06(mol)$
$C\%_{HCl} = \dfrac{0,06.36,5}{200}.100\% = 1,095\%$

b)

$m_{dd} = 3,04 + 200 = 203,4(gam)$
$C\%_{NaCl} = \dfrac{0,02.58,5}{203,4}.100\% = 0,58\%$

$C\%_{KCl} =\dfrac{0,04.74,5}{203,4}.100\% = 1,47\%$

Zn+2Hcl->Zncl2+H2

3----6--------3--------3

n Zn=\(\dfrac{195}{65}\)=3 mol

=>VH2=3.22,4=67,2l

=>mHCl=3.36,5=109,5g

Zn+2Hcl->Zncl2+H2

3-----6------3------ 3

n_Zn=

Ta có: \(m_{NaOH}=200.20\%=40\left(g\right)\Rightarrow n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

a, Theo PT: \(n_{HCl}=n_{NaOH}=1\left(mol\right)\)

\(\Rightarrow C\%_{HCl}=\dfrac{0,1.36,5}{100}.100\%=36,5\%\)

b, \(n_{NaCl}=n_{NaOH}=1\left(mol\right)\)

Ta có: m _{dd sau pư} = 200 + 100 = 300 (g)

\(\Rightarrow C\%_{NaCl}=\dfrac{1.58,5}{300}.100\%=19,5\%\)

a)

$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$

b)

$n_{CO_2} = n_{CaCO_3} = \dfrac{20}{100} = 0,2(mol)$

$V_{CO_2} = 0,2.22,4 = 4,48(lít)$

c)

Sau phản ứng : 

$m_{dd} = 20 + 200 - 0,2.44 = 211,2(gam)$

$n_{CaCl_2} = n_{CaCO_3} = 0,2(mol)$

$C\%_{CaCl_2} = \dfrac{0,2.111}{211,2}.100\% = 10,51\%$

a, PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

b, \(n_{CO_2}=n_{CaCl_2}=n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\text{​​}\right)\)

\(\Rightarrow V_{CO_2}=0,24,48\left(l\right)\)

c, \(m_{ddsaupu}=200+20-44.0,2=211,2\left(g\right)\)

\(\Rightarrow C\%=\dfrac{0,2.111}{211,2}.100\%=10,51\%\)

\(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\\ a.CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ b.n_{CO_2}=n_{CaCl_2}=n_{CaCO_3}=0,2\left(mol\right)\\ m_{ddCaCl_2}=20+200-0,2.44=211,2\left(g\right)\\ V_{CO_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c.m_{CaCl_2}=0,2.111=22,2\left(g\right)\\ C\%_{ddCaCl_2}=\dfrac{22,2}{211,2}.100\approx10,511\%\)

\(m_{Fe_2\left(SO_4\right)_3}=\dfrac{200\cdot20}{100}=40\left(g\right)\Rightarrow n=0,1mol\)

\(Fe_2\left(SO_4\right)_3+6NaOH\rightarrow2Fe\left(OH\right)_3\downarrow+3Na_2SO_4\)

0,1                    0,6              0,2                   0,3

a)\(m_{NaOH}=0,6\cdot40=24\left(g\right)\)

b)\(m_{Fe\left(OH\right)_3}=0,2\cdot107=21,4\left(g\right)\)

c)\(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)

   \(m_{ddsau}=200+24-21,4=202,6\left(g\right)\)

   \(\Rightarrow C\%=\dfrac{42,6}{202,6}\cdot100\%=21,03\%\)

\(a,n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow n_{HCl}=0,4\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,4\cdot36,5=14,6\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{14,6}{200}\cdot100\%=7,3\%\\ b,n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V_{H_2\left(đkc\right)}=0,2\cdot24,79=4,958\left(l\right)\\ c,m_{H_2}=0,2\cdot2=0,4\left(g\right)\\ n_{FeCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{CT_{FeCl_2}}=0,2\cdot127=25,4\left(g\right)\\ \Rightarrow m_{dd_{FeCl_2}}=11,2+200-0,4=210,8\left(g\right)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{25,4}{210,8}\cdot100\%\approx12,05\%\)