\(n_{Al}=\dfrac{8.1}{27}=0.3\left(mol\right)\)

\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)

\(0.3.....0.225........0.15\)

\(m_{Al_2O_3}=0.15\cdot102=15.3\left(g\right)\)

\(V_{O_2}=0.225\cdot22.4=5.04\left(l\right)\)

\(n_{CH_4}=\dfrac{3.2}{16}=0.2\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{^{^{t^0}}}CO_2+2H_2O\)

\(0.2......0.4........0.2......0.4\)

\(m_{H_2O}=0.4\cdot18=7.2\left(g\right)\)

\(V_{O_2}=0.4\cdot22.4=8.96\left(l\right)\)

\(V_{CO_2}=0.2\cdot22.4=4.48\left(l\right)\)

Bài 27: Hòa tan hết 4,6 gam A (I) bằng dung dịch HCl vừa đủ, thu được 2,24 lit H2(đktc).

a. Xác định A.

b. Tính số phân tử HCl cần dùng.

c. Tính mACl thu được (theo ĐLBTKL).

---

nH2=0,1(mol)

PTHH: A +HCl -> ACl + 1/2 H2

0,2_______0,2___0,2________0,1(mol)

=> M(A)= mA/nA=4,6/0,2=23(g/mol)

=>A(I) cần tìm là Natri (Na=23)

b) Số phân tử HCl cần dùng:

0,2.6.10²³=1,2.10²³ (phân tử)

c) mACl= mHCl+ mA - mH2= 0,2.36,5 + 4,6 - 0,1.2=11,7(g)

nFe3O4 = 23.2/232 = 0.1 mol

3Fe + 2O2 -to-> Fe3O4 

0.3____0.2_______0.1

mFe = 0.3*56 = 16.8 g 

VO2 = 0.2*22.4 = 4.48 (l) 

Vkk = 5VO2 = 22.4 (l) 

\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

PTHH : \(4Fe+3O_2\rightarrow2Fe_2O_3\)

             0,2        0,15      0,1         (mol)

\(m_{Fe}=0,2.56=11,2\left(g\right)\)

\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

______0,1->0,075-->0,05

a) V_O2 = 0,075.22,4 = 1,68(l)

b) m_Al2O3 = 0,05.102 = 5,1 (g)

a.n_Al=2,7/27=0,1(mol)
PTHH: 4Al + 3O2 --t--> 2Al2O3
(mol)     4         3                2
(mol)     0,1     0,075          0,05
n_O2=0,1.3/4=0,075 mol
VO2 đã dùng (đktc) là: 0,075.22,4=1,68(l)
b.m_Al2O3 = 0,05.102 = 5,1 (g)

\(n_S=\dfrac{3.2}{32}=0.1\left(mol\right)\)

\(n_{O_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(S+O_2\underrightarrow{t^0}SO_2\)

\(0.05.0.05...0.05\)

\(\Rightarrow Sdư\)

\(V_{SO_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(b.\)

\(S+O_2\underrightarrow{t^0}SO_2\)

\(0.1..0.1\)

\(V_{kk}=5V_{O_2}=5\cdot0.1\cdot22.4=11.2\left(l\right)\)

a, PT: \(S+O_2\underrightarrow{t^o}SO_2\)

Ta có: \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\), ta được S dư.

Theo PT: \(n_{SO_2}=n_{O_2}=0,05\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,05.22,4=1,12\left(l\right)\)

b, Theo PT: \(n_{O_2}=n_S=0,1\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)

\(\Rightarrow V_{kk}=2,24.5=11,2\left(l\right)\)

a. PTHH: 4 A l + 3 O 2 t o → 2 A l 2 O 3 b. n A l = 21 , 6 27 = 0 , 8 ( m o l ) 4 A l + 3 O 2 t o → 2 A l 2 O 3 Theo pt: 4 3 2 Theo đề: 0,8 0,6 0,4 (mol) m A l 2 O 3 = 0 , 4 × 102 = 40 , 8 ( g a m ) c. V O 2 = 0 , 6 × 22 , 4 = 13 , 44 ( l í t )

\(n_{Al}=\dfrac{21,6}{27}=0,8mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

0,8      0,6             0,4           ( mol )

\(V_{O_2}=0,6.22,4=13,44l\)

\(m_{Al_2O_3}=0,4.102=40,8g\)

\(V_{kk}=13,44.5=67,2l\)

Bạn xem lời giải ở đây nhé.

https://hoc24.vn/cau-hoi/cho-324-g-al-tac-dung-voi-oxi-vua-du-th-duoc-al2o3-a-tinh-vo2-b-tinh-m-al2o3-c-trong-vkk-can-dung-biet-vo2-21-vkk-d-tinh-khoi-luong-kmno.7651142171785

\(n_{CH_4}=\dfrac{0.8}{16}=0.05\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)

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

\(V_{kk}=5V_{O_2}=5\cdot0.1\cdot22.4=11.2\left(l\right)\)

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

\(a.C_2H_5OH+3O_2-^{t^o}\rightarrow2CO_2+3H_2O\\ n_{C_2H_5OH}=0,3\left(mol\right)\\ n_{CO_2}=2n_{C_2H_5OH}=0,6\left(mol\right)\\ \Rightarrow V_{CO_2}=0,6.22,4=13,44\left(l\right)\\ b.n_{O_2}=3n_{C_2H_5OH}=0,6\left(mol\right)\\ MàV_{O_2}=\dfrac{1}{5}V_{kk}\\ \Rightarrow V_{kk}=V_{O_2}.5=0,6.22,4.5=67,2\left(l\right)\\ c.n_{NaOH}=0,9\left(mol\right)\\ Tacó:\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,9}{0,6}=1,5\\ \Rightarrow Tạora2muốiNaHCO_3vàNa_2CO_3\\ Đặt:n_{NaHCO_3}=x\left(mol\right);n_{Na_2CO_3}=y\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}x+y=0,6\left(BTnguyento\left(C\right)\right)\\x+2y=0,9\left(BTnguyento\left(Na\right)\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,3\end{matrix}\right.\\ \Rightarrow m_{muối}=0,3.84+0,3.106=57\left(g\right)\)

a)C2H6O+3O2to→2CO2+3H2O

0,3-------------0,9------0,6

nC2H6O=\(\dfrac{13,8}{46}\)=0,3mol

nCO2=2nC2H6O=0,6mol

⇒VCO2=0,6×22,4=13,44l

⇒Vkk=0,9×22,4×5=100,8l