Question

Answer 1

a. 4Al+3O2-to>2Al2O3

 Xét tỷ số: 

0.2/4<0.3/3

=> Al hết, O2 dư. Vậy tính theo nAl

nO2=\(\dfrac{3}{4}\)0.2=0.15mol

nO2 dư=0.3-0.15=0.15mol

=>mO2 dư=0.15.32=4.8gam

2,

Zn+2HCl->ZnCl2+H2

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

2H2+O2-to>2H2O

0,2----0,1

n Zn=0,2 mol

=>Vkk=0,1.22,4.5=11,2l

Answer 2

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ pthh:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) 
 \(LTL:\dfrac{0,2}{4}< \dfrac{0,3}{3}\) 
=> Oxi dư 
\(n_{O_2\left(p\text{ư}\right)}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\\ n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\\ m_{O_2\left(d\right)}=\left(0,3-0,15\right).32=4,8\left(g\right)\\ m_{Al_2O_3}=0,1.102=10,2\left(g\right)\\ m_{sau-p\text{ư}}=4,8+\left(0,1.102\right)=15\left(g\right)\) 
b) 
\(n_{Zn}:\dfrac{13}{65}=0,2\left(mol\right)\\ pthh:Zn+2HCl\rightarrow H_2+ZnCl_2\) 
           0,2                 0,2 
\(pthh:2H_2+O_2\underrightarrow{t^o}2H_2O\) 
           0,2      0,1 
\(V_{kk}=\left(0,1.22,4\right):20\%=11,2\left(l\right)\)

Answer 3

nO2(pư)=