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

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

Theo PTHH: \(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow m_{muối}=0,1.161=16,1\left(g\right)\)

b. \(n_{H_2thu.được}=n_{Zn}=0,1\left(mol\right)\)

\(H_2+\dfrac{1}{2}O_2\underrightarrow{t^o}H_2O\)

0,1    0,05

\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)

\(\Rightarrow V_{không.khí}=1,12.5=5,6\left(l\right)\)

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: \(m_{Mg}+m_{O_2}=m_{MgO}\)

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

\(PTHH:Zn+2HCl->ZnCl_2+H_2\)

ap dung DLBTKL ta co

\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)

\(=>m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}\\ =>m_{H_2}=13+14,6-27,2\\ =>m_{H_2}=0,4\left(g\right)\)

a)

$Zn + H_2SO_4 \to ZnSO_4 + H_2$
$n_{Zn} = n_{H_2} = \dfrac{10,08}{22,4} = 0,45(mol)$
$m_{Zn} = 0,45.65 = 29,25(gam)$
b)

$n_{H_2SO_4} = n_{H_2} = 0,45(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,45.98}{20\%} = 220,5(gam)$

a) Zn + H2SO4 -> ZnSO4+ H2

nH2= 0,45(mol)

=>nZn=nH2SO4=nH2=0,45(mol)

=>mZn=0,45.65=29,25(g)

b) mH2SO4=0,45.98=44,1(g)

=>mddH2SO4=44,1. 100/20=220,5(g)

a) Zn + H2SO4 -------> ZnSO4 + H2

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

\(n_{Zn}=n_{H_2}=0,45\left(mol\right)\)

=> \(m_{Zn}=0,45.65=29,25\left(g\right)\)

b) \(n_{H_2SO_4}=n_{H_2}=0,45\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,45.98=44,1\left(g\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{44,1}{20\%}=220,5\left(g\right)\)

a. \(n_{Zn}=\dfrac{2,6}{65}=0,04\left(mol\right)\)

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

- Mol theo PTHH : \(1:1:1:1\)

- Mol theo phản ứng : \(0,04\rightarrow0,04\rightarrow0,04\rightarrow0,04\)

\(\Rightarrow m_{ZnSO_4}=n_{ZnSO_4}.M_{ZnSO_4}=0,04.161=6,44\left(g\right)\)

b. Từ a. suy ra : \(V_{H_2}=n_{H_2}.22,4=0,04.22,4=0,896\left(l\right)\)

c. Từ a. suy ra : \(n_{H_2}=0,04\left(mol\right)\)

\(PTHH:H_2+PbO\underrightarrow{t^o}Pb+H_2O\)

- Mol theo PTHH : \(1:1:1:1\)

- Mol theo phản ứng : \(0,04\rightarrow0,04\rightarrow0,04\rightarrow0,04\)

\(\Rightarrow m_{Pb}=n_{Pb}.M_{Pb}=0,04.207=8,28\left(g\right)\)

nZn = 19,5 : 65= 0,3 (mol) 
pthh Zn + 2HCl ---> ZnCl2 + H2 
       0,3--------------> 0,3-------> 0,3 (mol) 
=> mZnSO4 = 0,3 . 161 ( g) 
=> VH2 = 0,3 . 22,4 = 6,72 (l) 
nCuO = 16 : 80 =0,2 (mol) 
pthh : CuO + H2 -t--> Cu + H2O 
 LTL : 
0,2/1     < 0,3/1
=> H2 du 
ta co : nH2 (pu ) = nCuO = 0,2 (MOL) 
=> nH2(d)  = nH2 ( bd ) - nH2 (pu) = 0,3-0,2 = 0,1 (mol) 

nZn = 19,5/65 = 0,3 (mol)

PTHH: Zn + H2SO4 -> ZnSO4 + H2

Mol: 0,3 ---> 0,3 ---> 0,3 ---> 0,3

mZnSO4 = 0,3 . 161 = 48,3 (g)

VH2 = 0,3 . 22,4 = 6,72 (l)

nCuO = 16/80 = 0,2 (mol)

PTHH: CuO + H2 -> (t°) Cu + H2O

LTL: 0,2 < 0,3 => H2 dư

nH2 (pư) = 0,2 (mol)

mH2 (dư) = (0,3 - 0,2) . 2 = 0,2 (g)