a)\(PTHH:Zn+H_2SO_4\underrightarrow{ }ZnSO_4+H_2\)

b)\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(m\right)\);\(n_{H_2SO_4}=\dfrac{1,57}{98}=0,16\left(m\right)\)

\(PTHH:Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\)

ta có tỉ lệ:\(\dfrac{0,3}{1}>\dfrac{0,16}{1}->Zndư\)

\(n_{Zn\left(dư\right)}=0,3-0,16=0,14\left(m\right)\)

\(m_{Zn\left(dư\right)}=0,14.65=9,1\left(g\right)\)

c)\(PTHH:Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\)

tỉ lệ        :1       1                1              1

số mol   :0,16   0,16           0,16         0,16

\(V_{H_2}=0,16.22,4=3,584\left(l\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{1,95}{65}=0,03\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{1,57}{98}\approx0,016\left(mol\right)\)

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

\(\dfrac{n_{Zn\left(ĐB\right)}}{n_{Zn\left(PT\right)}}=\dfrac{0,03}{1}>\dfrac{n_{H_2SO_4\left(ĐB\right)}}{n_{H_2SO_4}\left(PT\right)}=\dfrac{0,016}{1}\)

\(\Rightarrow\) Zn dư , H₂SO₄ hết , tính theo H₂SO₄  

b, Theo PT : \(n_{zn}=n_{H_2SO_4}=0,016\left(mol\right)\)

\(\Rightarrow m_{Zn\left(pứ\right)}=n\cdot M=0,016\cdot32=0,512\left(g\right)\)

\(\Rightarrow m_{Zn\left(dư\right)}=m_{Zn\left(ĐB\right)}-n_{Zn\left(Pứ\right)}=1,95-0,512=1,438\left(g\right)\)

c, Theo PT : \(n_{H_2}=n_{H_2SO_4}=0,016\left(mol\right)\)

\(\Rightarrow V_{H_{2\left(đktc\right)}}=n\cdot22,4=0,016\cdot22,4=0,3584\left(l\right)\)

\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\\ m_{H_2SO_4}=\dfrac{22,05.20}{100}=4,41\left(g\right)\\ n_{H_2SO_4}=\dfrac{4,41}{98}=0,045\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\) 
\(LTL:\dfrac{0,03}{1}< \dfrac{0,045}{1}\) 
=> H₂SO₄ dư 
\(n_{H_2SO_4\left(p\text{ư}\right)}=n_{ZnSO_4}=n_{H_2}=n_{Zn}=0,03\left(mol\right)\\ m_{H_2SO_4\left(d\right)}=\left(0,045-0,03\right).98=1,47\left(g\right)\\ m_{\text{dd}}=1,95+22,05-\left(0,03.2\right)=23,94\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,03.136}{23,94}.100\%=17\%\)

\(a,n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\\ n_{H_2SO_4}=\dfrac{22,05}{98}=0,225\left(mol\right)\)

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

bđ       0,03   0,225

pư       0,03    0,03 

spư     0         0,195       0,03        0,03

\(b,m_{H_2SO_4\left(dư\right)}=0,195.98=19,11\left(g\right)\\ c,m_{dd}=1,95+22,05-0,03.2=23,94\left(g\right)\\ C\%_{ZnSO_4}=\dfrac{0,03.161}{23,94}.100\%=20,18\%\)

a) Zn + 2HCl --> ZnCl₂ + H₂

Phản ứng thế

b) \(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

_____0,02------------>0,02-->0,02

=> m_ZnCl2 = 0,02.136 = 2,72(g)

=> V_H2 = 0,02.22,4 = 0,448(l)

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

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

Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, \(n_{Fe_3O_4}=\dfrac{18,56}{232}=0,08\left(mol\right)\)

PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

Xét tỉ lệ: \(\dfrac{0,08}{1}>\dfrac{0,2}{4}\), ta được Fe₃O₄ dư.

Theo PT: \(n_{Fe}=\dfrac{3}{4}n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{16,25}{65}=0,25\left(mol\right)\\ PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)

tỉ lệ        1     :     1       :       1            :  1

n(mol)     0,25-->0,25------->0,25------>0,25

\(V_{H_2\left(dktc\right)}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\\ m_{ZnSO_4}=n\cdot M=0,25\cdot\left(65+32+16\cdot4\right)=40,25\left(g\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)

a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

b, Ta có: \(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{1,47}{98}=0,015\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,03}{1}>\dfrac{0,015}{1}\), ta được Zn dư.

Theo PT: \(n_{Zn\left(pư\right)}=n_{H_2SO_4}=0,015\left(mol\right)\Rightarrow n_{Zn\left(dư\right)}=0,03-0,015=0,015\left(mol\right)\)

\(\Rightarrow m_{Zn\left(dư\right)}=0,015.65=0,975\left(g\right)\)

c, \(n_{H_2}=n_{H_2SO_4}=0,015\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,015.22,4=0,336\left(l\right)\)

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\)

Phản ứng thế

\(b,n_{Zn}=\dfrac{1,3}{65}=0,02(mol)\\ \Rightarrow n_{ZnCl_2}=n_{H_2}=0,02(mol)\\ \Rightarrow m_{ZnCl_2}=0,02.136=2,72(g)\\ V_{H_2}=0,02.22,4=0,448(l)\)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ LTL:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCldư\\ n_{HCl\left(pứ\right)}=2n_{Zn}=0,4\left(mol\right)\\\Rightarrow m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65\left(g\right)\\ b.n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\ c.n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4,=4,48\left(l\right)\\ d.3H_2+Fe_2O_3-^{t^o}\rightarrow2Fe+3H_2O \\ n_{Fe_2O_3}=\dfrac{19,2}{160}=0,12\left(mol\right)\\ LTL:\dfrac{0,2}{3}< \dfrac{0,12}{1}\Rightarrow Fe_2O_3dưsauphảnứng\\ \Rightarrow n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}\left(mol\right)\\ \Rightarrow m_{Fe}=\dfrac{2}{15}.56=7,467\left(g\right)\)

a) n\(Zn\)=\(\dfrac{m}{M}\)=\(\dfrac{13}{65}\)=0,2(mol)

n\(HCl\)=\(\dfrac{m}{M}\)=\(\dfrac{18,25}{36,5}=\)0,5(mol)

PTHH : Zn + 2HCl->ZnCl\(2\) + H\(2\)

            0,2     0,5

Lập tỉ lệ mol : \(^{\dfrac{0,2}{1}}\)<\(\dfrac{0,5}{2}\)

n\(Zn\) hết , n\(HCl\) dư

-->Tính theo số mol hết

               Zn + 2HCl->ZnCl\(2\) + H\(2\)

              0,2 ->  0,4      0,2        0,2

n\(HCl\) dư= n\(HCl\)(đề) - n\(HCl\)(pt)= 0,5 - 0,4 = 0,1(mol)

m​\(HCl\) dư= 0,1.36,5 = 3,65(g)

b) m\(ZnCl2\) = n.M= 0,2.136= 27,2 (g)

c)V\(H2\)=n.22,4=0,2.22,4=4,48(l)

d) n\(Fe\)\(2\)O\(3\)=\(\dfrac{m}{M}\)=\(\dfrac{19,2}{160}\)=0,12 (mol)

 3H2 +Fe2O3 → 2Fe + 3H2O

  0,2       0,12

Lập tỉ lệ mol: \(\dfrac{0,2}{3}\)<\(\dfrac{0,12}{1}\)

    nH2 hết .Tính theo số mol hết

\(HCl\)

​ 3H2 +Fe2O3 → 2Fe + 3H2O

0,2->                  0,2

m\(Fe\)=n.M= 0,2.56= 11,2(g)