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

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

Theo PT: \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

c, \(n_{ZnSO_4}=n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ZnSO_4}=0,3.161=48,3\left(g\right)\)

d, \(n_{Fe_2O_3}=\dfrac{64}{160}=0,4\left(mol\right)\)

PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,3}{3}\), ta được Fe₂O₃ dư.

Mà: H% = 30% \(\Rightarrow n_{H_2\left(pư\right)}=0,3.30\%=0,09\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{2}{3}n_{H_2}=0,06\left(mol\right)\\n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=0,03\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{Fe_2O_3\left(dư\right)}=0,4-0,03=0,37\left(mol\right)\)

\(\Rightarrow a=m_{Fe}+m_{Fe_2O_3\left(dư\right)}=62,56\left(g\right)\)

\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ b) n_{H_2} = n_{Zn} = \dfrac{13}{65}=0,2(mol)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\\ c) CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{CuO} = n_{H_2} = 0,2(mol)\\ m_{CuO} = 0,2.80 = 16(gam)\)

a)

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

a,PTHH: Zn+H2SO4-----ZnSO4 +H2

nZn = 0,2 mol

=> VH2=4,48 l

CuO + H2 ----- Cu +H2O

nH2= 0,2 mol 

=> mCuO= 16g 

a) Zn + 2HCl →ZnCl₂  + H₂

b) nZn = 6,5/65 = 0,1 mol . Theo tỉ lệ pư => nH₂ = nZn = nZnCl₂ =0,1 mol <=> V_H2(đktc) = 0,1.22,4 = 2,24 lít.

c) mZnCl₂ = 0,1 . 136 = 13,6 gam

d) nHCl =2nZn = 0,2 mol => mHCl = 0,2.36,5= 7,3 gam

Cách 2: áp dụng định luật BTKL => mHCl = mZnCl₂ + mH₂ - mZn 

<=> mHCl = 13,6 + 0,1.2 - 6,5 = 7,3 gam

\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{19,6}{2+32+16\cdot4}=0,2\left(mol\right)\\ PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)

tỉ lệ:        1     :     1        :      1        :      1

n(mol)     0,2<---0,2------>0,2-------->0,2

\(m_{ZnSO_4}=n\cdot M=0,2\cdot\left(65+32+16\cdot4\right)=32,2\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\)

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

tỉ lệ        1     :     1       :       1            :  1

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

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

Zn  +  2HCl  → ZnCl₂  +   H₂

nMg = 13:65 = 0,2 mol

Theo tỉ lệ phương trình phản ứng => nH₂ = nZn = 0,2 mol

<=> VH₂ đktc = 0,2.22,4 = 4,48 lít

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

\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

             0,25      0,5       0,5        0,5

Xét tỉ lệ : \(\dfrac{0,3}{1}>\dfrac{0,5}{2}\) => Zn dư , HCl đủ 

b) \(m_{Zn\left(dư\right)}=\left(0,3-0,25\right).65=3,25\left(g\right)\)

c) \(m_{ZnCl_2}=0,25.136=34\left(g\right)\)

\(V_{H_2}=0,25.22,4=5,6\left(l\right)\)

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ a,Zn+2HCl\rightarrow ZnCl_2+H_2\\b, Vì:\dfrac{0,5}{2}< \dfrac{0,3}{1}\Rightarrow Zndư\\ n_{Zn\left(dư\right)}=0,3-\dfrac{0,5}{2}=0,05\left(mol\right)\\ \Rightarrow m_{Zn\left(dư\right)}=0,05.65=3,25\left(g\right)\\ c,n_{ZnCl_2}=n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,25.136=34\left(g\right)\\ V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

a) PTHH

Zn + 2HCl -> ZnCl₂ + H_2↑

0,25    0,5        0,25         0,25

n_Zn= \(\dfrac{19,6}{65}\)= 0,3(mol)

n_HCl= \(\dfrac{18,25}{36,5}\)=0,5 (mol)

So sánh nZn và nHCl 

            0,3/1  >  0,5/2

b) =>Zn tác dụng dư; HCl tác dụng hết

n_{Zn dư}= 0,3-0,25=0,05(mol)

=> m_{Zn dư}= 0,05 x 65= 3,25(mol)

c) m_ZnCl2= 0,25 x 136=34(g)

V_H2= 0,25 x 22,4= 5,6 (l)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\b.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\\ c.n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\\ m_{FeCl_2}=0,1.127=12,7\left(g\right) \)

\(a)Zn+2HCl\rightarrow ZnCl_2+H_2\\ b)n_{Zn}=\dfrac{13}{65}=0,2mol\\ n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,2mol\\ V_{H_2}=0,2.22,4=4,48l\\ c)n_{HCl}=0,2.2=0,4mol\\ C_{M_{HCl}}=\dfrac{0,4}{0,2}=2M\\ d)m_{ZnCl_2}=0,2.136=27,2g\)