a/Vo2=5,6l

b/mkclo3=20,4

Bạn làm lời giải chi tiết giúp mk nhé

Cảm ơn bạn ạ

Bạn làm như thế mình ko hiểu ạ

Zn + 1/2O2 ---> ZnO

nO2 = 1/2nZnO = 40,5/81/2 = 0,25 mol ---> V = 0,25.22,5 = 5,6 lít.

b) KClO3 ---> KCl + 3/2O2

nKClO3 = 2/3nO2 = 0,5/3 mol ---> m = 122,5.0,5/3 = 20,42 gam.

a)Pt 2Zn + O₂ \(\underrightarrow{t^o}\) 2 ZnO

-ⁿZnO=\(\dfrac{40.5}{81}=0.5\left(mol\right)\)

-Theo pt thì ⁿO₂=\(\dfrac{1}{2}\) ⁿZnO=0.25(mol)

=>^VO₂=0,25.22.4=5.6(l)

b) Pt : 2KClO₃\(\underrightarrow{t^o}\) 2KCl + 3O₂

-Theo phương trình : ⁿKClO₃=\(\dfrac{2}{3}\) ⁿO₂=\(\dfrac{1}{6}\left(mol\right)\)

=>^mKClO₃=1/6.122.5=24.42(g)

Vậy...

2Zn+O2--->2ZnO

a) n Zn=40,5/81=0,5(mol)

Theo pthh

n O2=1/2 n ZnO=0,25(mol)

V O2=0,25.22,4=5,6(l)

b) 2KClO3--->2KCl+3O2

Theo pthh

n KClO3=2/3n O2=0,16667(mol)

m KClO3=0,16667.122,5=20,42(g)

Chúc bạn học tốt

a, PTHH ( I ) : \(2Zn+O_2\rightarrow2ZnO\)

\(n_{ZnO}=\frac{m_{ZnO}}{M_{ZnO}}=\frac{40,5}{65+16}=0,5\left(mol\right)\)

- Theo PTHH ( I ) : \(n_{O_2}=\frac{1}{2}n_{ZnO}=\frac{1}{2}0,5=0,25\left(mol\right)\)

-> \(V_{O_2}=n_{O_2}.22,4=0,25.22,4=5,6\left(l\right)\)

b, PTHH ( II ) : \(2KClO_3\rightarrow2KCl+3O_2\)

- Theo PTHH ( II ) : \(n_{KClO_3}=\frac{2}{3}n_{O_2}=\frac{2}{3}0,25=\frac{1}{6}\left(mol\right)\)

-> \(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=\frac{1}{6}.122,5\approx20,4\left(g\right)\)

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

\(n_{ZnO}=\frac{40,5}{81}=0,5mol\)

Theo phương trình thì \(n_{O_2}=\frac{1}{2}n_{ZnO}=\frac{1}{2}.0,5=0,25mol\)

\(\Rightarrow V_{O_2}=0,25.22,4=5,6l\)

b)\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Theo phương trình thì \(n_{KClO_3}=\frac{2}{3}n_{O_2}=\frac{2}{3}.0,25=\frac{1}{6}mol\)

\(\Rightarrow m_{KClO_3}=\frac{1}{6}.122,5=24,42g\)

\(pthh:2Zn+O_2\overset{t^0}{\rightarrow}2ZnO\left(1\right)\)

\(n_{ZnO}=\dfrac{m}{M}=\dfrac{40,5}{81}0,5\left(mol\right)\)

Theo \(pthh\left(1\right):n_{O_2}=\dfrac{1}{2}n_{ZnO}=\dfrac{1}{2}\cdot0,5=0,25\left(mol\right)\)

\(\Rightarrow m_{O_2}=n\cdot M=0,25\cdot32=8\left(g\right)\)

\(2KClO_3\overset{t^0}{\rightarrow}2KCl+3O_2\left(2\right)\)

Theo \(pthh\left(2\right):n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}\cdot0,25=\dfrac{1}{3}\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=n\cdot M=\dfrac{1}{3}\cdot122,5=40,83\left(g\right)\)

\(n_{ZnO}=\dfrac{40.5}{81}=0,5\left(mol\right)\)

PTHH : 2Zn + O₂ -> 2ZnO

             0,5                0,5

\(m_{Zn}=0,5.65=32,5\left(g\right)\)

\(m_{bụi.kẽm}=\dfrac{32,5}{100\%-2\%}=33,16\left(g\right)\)

Chọn B

B

cảm ơn ạa

\(m_{ZnO}=40,5\left(g\right)\)

\(\rightarrow n_{ZnO}=\dfrac{40,5}{81}=0,5\left(mol\right)\)

\(2Zn+O_2\rightarrow2ZnO\)

0,5 ................. 0,5 (mol) 

\(m_{Zn}=0,5.65=32,5\left(g\right)\)

-> m bụi kẽm \(=\dfrac{32,5}{100\%-2\%}\approx33,16\left(g\right)\)

Nếu phản ứng có hiệu suất 100% thì

2Zn+O2−>2ZnO2Zn+O2−>2ZnO

0,5       0,25         0,5 mol

Khối lượng kẽm cần dùng là: 65*0.5= 32,5 (g)

Khối lượng oxi cần dùng là: 40,5-32,5= 8 (g)

Bài 1

a)\(2Zn+O2-->2ZnO\)

b)\(n_{ZnO}=\frac{48,6}{81}=0,6\left(mol\right)\)

\(n_{O2}=\frac{1}{2}n_{Zn}=0,3\left(mol\right)\)

\(V_{O2}=0,3.22,4=6,72\left(l\right)\)

c)Cách 1

\(n_{Zn}=n_{ZnO}=0,6\left(mol\right)\)

\(m_{Zn}=0,6.65=39\left(g\right)\)

Cách 2

\(m_{O2}=0,3.32=9,6\left(g\right)\)

\(m_{Zn}=m_{ZnO}-m_{O2}=48,6-9,6=39\left(g\right)\)

Bài 2

a)\(m_{Mg}=9,6.12,5\%=1,2\left(g\right)\)

\(m_{Fe}=9,6-1,2=8,4\left(g\right)\)

b)\(2Mg+O2-->2MgO\)

\(3Fe+2O2-->Fe3O4\)

\(n_{Mg}=\frac{1,2}{24}=0,05\left(mol\right)\)

\(n_{Fe}=\frac{8,4}{56}=0,15\left(mol\right)\)

\(n_{MgO}=n_{Mg}=0,05\left(mol\right)\)

\(m_{gO}=0,05.40=2\left(g\right)\)

\(n_{Fe3O4}=\frac{1}{3}n_{Fe}=0,05\left(mol\right)\)

\(m_{Fe3O4}=0,05.232=11,6\left(g\right)\)