\(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

0,1_______0,1____________0,1______________

\(m_{muoi}=0,1.\left(12+3+12+16+16\right).2+65=18,3\left(g\right)\)

\(\Rightarrow V_{H2}=0,1.22,4=2,24\left(l\right)\)

Zn+CH3COOH--->H2+(CH3COO)2Zn

n Zn=6,5/65=0,1(mol)

n H2=n Zn=0,1(mol)

V H2=0,1.22,4=2,24(l)

n (CH3COO)2Zn=n Zn=0,1(mol)

m (CH3COO)2Zn=0,1.183=18,3(g)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\\ m_{muối}=m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\ V_{khí\left(đktc\right)}=V_{H_2\left(đkc\right)}=0,1.24,79=2,479\left(l\right)\\ c,n_{CuO}=\dfrac{7,6}{80}=0,095\left(mol\right)\\ PTHH:CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Vì:\dfrac{0,095}{1}< \dfrac{0,1}{1}\Rightarrow H_2dư\\ n_{Cu}=n_{CuO}=0,095\left(mol\right)\\ m_{Cu}=0,095.64=6,08\left(g\right)\)

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

PTHH :

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

0,3      0,6         0,3         0,3 

\(a,V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)

\(m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

\(b,V_{ddHCl}=\dfrac{n}{C_M}=\dfrac{0,6}{2}=0,3\left(l\right)\)

\(c,Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

     0,1         0,3 

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Ta có : 

\(\dfrac{0,1}{1}=\dfrac{0,3}{3}\)

nên không chất nào dư 

a, \(m_{CH_3COOH}=100.6\%=6\left(g\right)\Rightarrow n_{CH_3COOH}=\dfrac{6}{60}=0,1\left(mol\right)\)

PT: \(CH_3COOH+NaHCO_3\rightarrow CH_3COONa+CO_2+H_2O\)

Theo PT: \(n_{NaHCO_3}=n_{CH_3COONa}=n_{CO_2}=n_{CH_3COOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{NaHCO_3}=0,1.84=8,4\left(g\right)\)

\(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

b, Ta có: m _{dd sau pư} = 100 + 8,4 - 0,1.44 = 104 (g)

\(\Rightarrow C\%_{CH_3COONa}=\dfrac{0,1.82}{104}.100\%\approx7,88\%\)

\(a,n_{\left(CH_3COO\right)_2Mg}=\dfrac{7,1}{142}=0,05\left(mol\right)\)

PTHH: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)

                        0,1<----------------0,05-------------->0,05

\(\rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\\ b,C\%_{CH_3COOH}=\dfrac{0,1.60}{100}.100\%=6\%\)

\(c,n_{C_2H_5OH}=\dfrac{6,9}{46}=0,15\left(mol\right)\)

PTHH: \(CH_3COOH+C_2H_5OH\xrightarrow[H_2SO_{4\left(đặc\right)}]{t^o}CH_3COOC_2H_5+H_2O\)

bđ          0,1                 0,15

pư          0,1                 0,1

spư         0                     0,05                          0,1

\(\rightarrow m_{este}=0,1.80\%.88=7,04\left(g\right)\)

\(n_{HCl}=0,25.2=0,5\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{Zn}=0,25.65=16,25\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

Không biết đúng không nữa;-;;;

a) PTHH: Zn + 2HCl -> ZnCl2 + H2 

b) HCl=250ml=0,25l

n2HCl= V/22,4= 0,5/22,4= 0,02(mol)

Zn  +  2HCl -> ZnCl2  +  H2

1          2              1         1

0,01 <-0,5--------------> 0,01

mZn= n.M= 0,01.65= 0,65(gam)

c) VH2=n . 22,4= 0,01 . 22,4= 0,224(l)
 

\(a,PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ b,n_{FeCl_2}=n_{H_2}=n_{Fe}=0,4\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ c,m_{FeCl_2}=127.0,4=50,8\left(g\right)\)

\(n_{HCl}=0,3.2=0,6\left(mol\right)\\a, 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b,n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{Al}=n_{AlCl_3}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ m_{Al}=0,2.27=5,4\left(g\right)\\ d,V_{ddAlCl_3}=V_{ddHCl}=0,3\left(l\right)\\ C_{MddHCl}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)

a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

            0,1      0,2                  0,1

b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

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

\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)

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

0,1      0,2                       0,1

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{HCl}=0,2\cdot36,5=7,3g\)