Phương trình hóa học :

\(2Zn + O_2 \xrightarrow{t^o} 2ZnO\)

Ta có : \(n_{Zn} = \dfrac{1,3}{65} = 0,02(mol)\)

Theo PTHH :

\(n_{ZnO} = n_{Zn}=0,02(mol)\\ \Rightarrow m_{ZnO} = 0,02.81 = 1,62(gam)\)

Phương trình hóa học:

\(2Zn+O_2\dfrac{^{t^o}}{^{\rightarrow}}\)\(2ZnO\)

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

Theo PTHH:

\(n_{ZnO}=n_{Zn}=0,02\left(mol\right)\)

\(\Rightarrow m_{ZnO}=0,02.81=1,62\left(g\right)\)

\(a,PTHH:2Zn+O_2\rightarrow^{t^o}2ZnO\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ \Rightarrow n_{O_2}=\dfrac{1}{2}n_{Zn}=0,1\left(mol\right)\\ \Rightarrow m_{O_2}=0,1\cdot32=3,2\left(g\right)\\ c,\text{Bảo toàn KL: }m_{ZnO}=m_{O_2}+m_{Zn}=3,2+13=16,2\left(g\right)\)

PTHH: \(Zn+\dfrac{1}{2}O_2\xrightarrow[]{t^o}ZnO\)

Bảo toàn khối lượng: \(m_{O_2}=m_{ZnO}-m_{Zn}=1,6\left(g\right)\)

a. \(2Zn+O_2\rightarrow2ZnO\)

b.\(m_{Zn}+m_{O_2}\rightarrow m_{ZnO}\)

\(\Rightarrow6,5+m_{O_2}=8,1\)

\(\Rightarrow m_O=8,1-6,5=1,6\)

\(2Zn+O_2\underrightarrow{^{^{t^0}}}2ZnO\)

Bảo toàn khối lượng : 

\(m_{Zn}+m_{O_2}=m_{ZnO}\)

\(\Rightarrow m_{O_2}=m_{ZnO}-m_{Zn}=20-10.5=9.5\left(g\right)\)

\(2Zn+O_2\underrightarrow{^{^{t^0}}}2ZnO\)

Bảo toàn khối lượng : 

\(m_{Zn}+m_{O_2}=m_{ZnO}\)

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

PTHH: 2Zn + O2 -> (t°) 2ZnO

Mol: 0,2 ---> 0,1 ---> 0,2

b, VO2 = 0,1 . 22,4 = 2,24 (l)

c, mZnO = 0,2 . 81 = 16,2 (g)

d, PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2

nKMnO4 = 0,1 . 2 = 0,2 (mol)

mKMnO4 = 0,2 . 158 = 31,6 (g)

2Zn+O2-to>2ZnO

0,2---0,1-----0,2

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

=>VO2=0,1.22,4=2,24l

=>m ZnO=0,2.81=16,2g

d)

2KMnO4-to>K2MnO4+MnO2+O2

0,2-----------------------------------0,1

=>m KMnO4=0,2.158=31,6g

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

PTHH: 2Zn + O2 -> (t°) 2ZnO

Mol: 0,2 ---> 0,1 ---> 0,2

b, VO2 = 0,1 . 22,4 = 2,24 (l)

c, mZnO = 0,2 . 81 = 16,2 (g)

d, PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2

nKMnO4 = 0,1 . 2 = 0,2 (mol)

mKMnO4 = 0,2 . 158 = 31,6 (g)

a) 

$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$2Zn + O_2 \xrightarrow{t^o} 2ZnO$

b) Theo ĐLBTKL:

\(m_{Kim.lo\text{ại}}+m_{O_2}=m_{\text{ox}it}\)

=> \(m_{O_2}=13,7-11=2,7\left(g\right)\)

\(2Zn + O_2 \xrightarrow{t^o} 2ZnO\)

Bảo toàn khối lượng :

\(m_{Zn} + m_{O_2} = m_{ZnO}\\ \Rightarrow m_{O_2} = 8,1 - 6,5=1,6 (gam)\)

a) PTHH : \(2Zn+O_2-t^o->2ZnO\)

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

Theo PTHH : \(n_{O2}=\dfrac{1}{2}n_{Zn}=0,15\left(mol\right)\)

=> \(V_{O2}=0,15.22,4=3,36\left(l\right)\)

c) Theo PTHH : \(n_{ZnO}=n_{Zn}=0,3\left(mol\right)\)

=> \(m_{ZnO}=0,3.81=24,3\left(g\right)\)

vậy ...

\(\begin{array}{l} a,\ PTHH:2Zn+O_2\xrightarrow{t^o} 2ZnO\\ b,\\ n_{Zn}=\dfrac{19,5}{65}=0,3\ (mol)\\ Theo\ pt:\ n_{O_2}=\dfrac{1}{2}n_{Zn}=0,15\ (mol)\\ \Rightarrow V_{O_2}=0,15\times 22,4=3,36\ (l)\\ c,\\ Theo\ pt:\ n_{ZnO}=n_{Zn}=0,3\ (mol)\\ \Rightarrow m_{ZnO}=0,3\times 81=24,3\ (g)\end{array}\)