Bài 22: Tính theo phương trình hóa học

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

a, PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

___0,1__0,05__0,1 (mol)

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

c, Ta có: \(m_{H_2o}=0,1.18=1,8\left(g\right)\)

Bạn tham khảo nhé!

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(2H_2+O_2\underrightarrow{t^0}2H_2O\)

\(0.1......0.05....0.1\)

\(V_{O_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(m_{H_2O}=0.1\cdot18=1.8\left(g\right)\)

a/ PTHH: \(2H_2+O_2\overset{t^o} 2H_2O\)

b/ \(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\)

Theo pt: \(2:1:2\)

\(\to n_{O_2}=0,05(mol)\)

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

b/ \(n_{H_2O}=0,1(mol)\)

\(\to m_{H_2O}=0,1.18=1,8(g)\)

\(n_{O_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)

\(.......0.13....0.075\)

\(V_{O_2}=3.36\left(l\right)\)

\(m_{Fe_3O_4}=0.075\cdot232=17.4\left(g\right)\)

a) pt: 3Fe + 2O₂ \(\rightarrow\) Fe₃O₄

b)  Thể tích khí oxi cho ở đề bài rồi mà

c) Theo pt: nFe₃O₄ = \(\dfrac{1}{2}n_{O_2}=\dfrac{1}{2}.0,15=0,075mol\)

\(\Rightarrow mFe_3O_4=0,075.232=17,4g\)

a) nKMnO4= 63,2/158=0,4(mol)

nFe=0,5(mol)

PTHH: KMnO4 -to-> K2MnO4 + MnO2 + 1/2 O2

0,4________________________________0,2(mol)

=>V(O2,đktc)=0,2.22,4=4,48(l)

PTHH: Fe + 2 HCl -> FeCl2 + H2

0,5_____________________0,5(mol)

=> V(H2,đktc)=0,5.22,4=11,2(l)

b) H2 + 1/2 O2 -to-> H2O

0,5/2 > 0,2/1

=> H2 dư, O2 hết, tính theo nO2

=> nH2(dư)=0,5 - 0,2.2=0,1(mol)

=>mH2(dư)=0,1.2=0,2(g)

c) nFe3O4=166/232= 83/116(mol)

PTHH: Fe3O4 + 4 H2 -to-> 3 Fe + 4 H2O

nFe3O4(p.ứ)=0,5/4=0,125(mol)

=> mFe3O4=0,125.232=29(g)

nFe=3/4. 0,5=0,375(mol)

=>mFe=0,375.56=21(g)

a) PTHH: NaOH + Al + H2O -> NaAlO2 + 3/2 H2

b) nH2= 0,6(mol)

-> nAl=0,4(mol) => mAl=0,4.27=10,8(g)

c) nAl=0,18((mol); nNaOH=0,2(mol)

PTHH: 0,18/1 < 0,2/1

=> Al hết, NaOH dư, tính theo nAl.

-> nH2= 3/2. 0,18=0,27(mol)

=>V(H2,đktc)=0,27.22,4= 6,048(l)

\(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)

\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)

\(...........0.4.........................0.6\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

\(n_{Al}=\dfrac{4.86}{27}=0.18\left(mol\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0.2\left(mol\right)\)

\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)

\(2.................2\)

\(0.2...............0.18\)

\(LTL:\dfrac{0.2}{2}>\dfrac{0.18}{2}\)

\(\Rightarrow NaOHdư\)

\(n_{H_2}=0.18\cdot\dfrac{3}{2}=0.27\left(mol\right)\)

\(V_{H_2}=0.27\cdot22.4=6.048\left(l\right)\)

a, PTHH: 2NaOH + 2Al + 2H₂O → 2NaAlO₂ + 3H₂↑

b, nH2=0,6 mol -mAl=10,8 g 

c, nAl=0,16(592) mol 

   nNaOH=0,2 mol 

-NaOH hết, Al dư 

-VH2=6,72 l

\(n_{H_2}=\dfrac{17.92}{22.4}=0.8\left(mol\right)\)

\(n_{Fe_3O_4}=\dfrac{69.6}{232}=0.3\left(mol\right)\)

\(Fe_3O_4+4H_2\underrightarrow{t^0}3Fe+4H_2O\)

\(0.2..............0.8\)

\(m_{Fe_3O_4\left(dư\right)}=\left(0.3-0.2\right)\cdot232=23.2\left(g\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.8......................................0.8\)

\(m_{Zn}=0.8\cdot65=52\left(g\right)\)

nP= 0,3(mol)

V(O2,dùng đktc)= Vkk/5= 26,88/5= 5,376(l)

->nO2=0,24(mol)

PTHH: 4P + 5 O2 -to-> 2 P2O5

Ta có: 0,3/4 > 0,24/5

=> P dư, O2 hết, tính theo nO2.

-> nP2O5= 2/5. nO2= 2/5 . 0,24=0,096(mol)

-> mP2O5=0,096.142=13,632(g)

nKClO3=0,2(mol)

nCu=0,5(mol)

PTHH: 2 KClO3 -to-> 2 KCl + 3 O2

0,2______________________0,3( mol)

Cu + 1/2 O2 -to-> CuO

Ta có: 0,5/2 < 0,3/1

=> O2 dư, Cu hết, tính theo nCu.

=> nCuO=nCu=0,5(mol)

=>mCuO=0,5.80=40(g)

nH2SO4= 0,125(mol)

nFe=0,2(mol)

PTHH: Fe + H2SO4 -> FeSO4 + H2

Vì: 0,125/1 < 0,2/1

=> Fe dư, H2SO4 hết, tính theo nH2SO4

-> nH2=nH2SO4=0,125(mol)

=>V(H2,đktc)=0,125.22,4=2,8(l)

a) nS= 0,1(mol)

nFe= 11/140(mol)

Vì: 0,1/1 < 11/140:1

=> Fe hết, S dư, tính theo nFe.

PTHH: Fe + S -to-> FeS

11/140____11/140__11/140(mol)

=> mFeS= 11/140 x 88 \(\approx\) 6,914(g)

\(a) n_P = \dfrac{18,6}{31} = 0,6(mol)\\ n_{O_2} = \dfrac{20,16}{22,4} = 0,9(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ \dfrac{n_P}{4} = 0,15 < \dfrac{n_{O_2}}{5} = 0,18 \to O_2\ dư\\ n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,75(mol)\\ \Rightarrow m_{O_2\ dư} = (0,9-0,75).32 = 4,8(gam)\\ b) n_{Fe} = \dfrac{56}{56} = 1(mol)\)

\(3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ \dfrac{n_{Fe}}{3} = \dfrac{1}{3}<\dfrac{n_{O_2}}{2} = 0,45\to Fe\ cháy\ hết.\\ c)\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,9.2 = 1,8(mol)\\ \Rightarrow m_{KMnO_4} = 1,8.158 =284,4(gam)\)