Zn + H2SO4 → ZnSO4 + H2

nZn=13/65=0,2 (mol)

nH2SO4=nZnSO4=nH2=0,2 (mol)

a) mH2SO4=0,2.98=19,6 (g)

b) mZnSO4=0,2.161=32,2 (g)

c) VH2=0,2.22,4=4,48 (l)

$PTHH:Zn+2HCl\to ZnCl_2+H_2\uparrow$

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

Theo PT: $n_{ZnCl_2}=n_{H_2}=0,2(mol);n_{HCl}=0,4(mol)$

$a)m_{axit}=m_{HCl}=n.M=0,4.36,5=14,6(g)$

$b)m_{ZnCl_2}=n.M=0,2.136=27,2(g)$

$c)V_{H_2(đktc)}=n.22,4=0,2.22,4=4,48(lít)$

Số mol kẽm là : 

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

PTHH : Zn + 2HCL -> ZnCl₂ + H₂

1 2 1 1 

0,2 mol -> 0,4 mol 0,2 mol 0,2 mol 

a, Khối lượng HCL là : 

\(m=n.M=0,4.35,5=14,2\left(g\right)\)

b, Khối lượng ZnCL₂ là : 

\(m=n.M=0,1.136=13,6\left(g\right)\)

c, Thể tích H₂ là : V = n . 22,4 = \(0,1.22,4=2,24\left(l\right)\)

a, Bảo toàn e:

\(2n_{Zn}=2n_{H_2}\)

\(\Leftrightarrow n_{Zn}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=19,5\left(g\right)\)

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

b, Bảo toàn nguyên tố H:

\(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=21,9\left(g\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)

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

b, Theo PT: \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)

Bài 1:

\(a,PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\\ b,m_{Zn}+m_{H_2SO_4}=m_{ZnSO_4}+m_{H_2}\\ c,m_{H_2SO_4}=32,2+0,4-13=19,6(g) \)

Bài 2:

 Bảo toàn KL: \(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)

\(\Rightarrow m_{H_2}=6,5+7,3-13,6=0,2(g)\)

Bài 3:

Bảo toàn KL: \(m_{Mg}+m_{O_2}=m_{MgO}\)

\(\Rightarrow m_{O_2}=1000-600=400(g)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1(mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{H_2}=0,1(mol);n_{HCl}=0,2(mol)\\ a,V_{H_2}=0,1.22,4=2,24(l)\\ b,m_{HCl}=0,2.36,5=7,3(g)\)

Bài1:

CuO: Cu(OH)2

FeO: Fe(OH)2

BaO: Ba(OH)2

MgO: Mg(OH)2

Bài 2

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

PTHH: Zn + H2SO4 -->ZnSO4 + H2

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

=> \(m_{H2SO4}=0,2.98=19,6\left(g\right)\)

b) \(m_{ZnSo4}=0,2.161=32,2\left(g\right)\)

c) \(V_{H2}=0,2.22,4=4,48\left(l\right)\)

Bài 3:

\(n_{CO2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)

a) PTHH: CaCO3 --to--> CaO + CO2

b) PTHH: CaCO3 --to--> CaO + CO2

0,25 <---------- 0,25 <-- 0,25 (mol)

=> \(m_{CaCO3}=0,25.100=25\left(g\right)\)

c) \(m_{CaO}=0,25.56=14\left(g\right)\)

\(Zn + H_2SO_4 \to ZnSO_4 + H_2\\ n_{Zn} = n_{ZnSO_4} = n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ m_{Zn} = 0,3.65 = 19,5(gam)\\ m_{ZnSO_4} = 0,3.161 = 48,3(gam)\)

a) Zn+H₂SO₄→ZnSO₄+H₂

b) Ta có : V_H2= 6,72(l) → n_H2= n_Zn=n_ZnSO4=\(\dfrac{6,72}{22,4}\)=0,3(mol)

m_Zn=0,3.65=19,5(gam)

m_ZnSO4=0,3.161=48,3(gam)