Bài 22. Tính theo phương trình hóa học

a)

\(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\)

\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)

Ta thấy : 

\(\dfrac{n_{Al}}{4} = 0,025 < \dfrac{n_{O_2}}{3} = 0,1\) nên O₂ dư

Theo PTHH : 

\(n_{Al_2O_3} = 0,5n_{Al} = 0,05(mol)\\ n_{O_2\ pư} = \dfrac{3}{4}n_{Al} = 0,075(mol)\)

Suy ra : 

\(m_{Al_2O_3} = 0,05.102 = 5,1(gam)\\ m_{O_2\ dư} = (0,3 - 0,075).32 = 7,2(gam)\)

PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

a) Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{4}< \dfrac{0,3}{3}\) \(\Rightarrow\) Oxi còn dư, Al phản ứng hết

\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2O_3}=0,05mol\\n_{O_2\left(dư\right)}=0,225mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=0,225\cdot32=7,2\left(g\right)\\m_{Al_2O_3}=0,05\cdot102=5,1\left(g\right)\end{matrix}\right.\)

b) Theo PTHH: \(n_{O_2\left(pư\right)}=0,075mol\)

\(\Rightarrow V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\)

Vì Oxi chiếm khoảng 20% thể tích không khí 

\(\Rightarrow V_{kk}=\dfrac{1,68}{20\%}=8,4\left(l\right)\)

\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)

\(n_{O_2}=0.3\left(mol\right)\)

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

\(0.1.....0.075.....0.05\)

\(m=m_{Al_2O_3}+m_{O_2\left(dư\right)}=0.05\cdot102+\left(0.3-0.075\right)\cdot32=12.3\left(g\right)\)

\(V_{kk}=5V_{O_2}=5\cdot0.075\cdot22.4=8.4\left(l\right)\)

Bài 1 : 

a)

\(2Cu + O_2 \xrightarrow{t^o} 2CuO\)

b)

Ta có :

\(n_{Cu} = \dfrac{32}{64} = 0,5(mol)\)

Theo PTHH : \(n_{O_2} = 0,5n_{Cu} = 0,25(mol)\\ \Rightarrow V_{O_2} = 0,25.22,4 = 5,6(lít)\)

c) Ta có : \(n_{CuO} = n_{Cu} = 0,5(mol)\Rightarrow m_{CuO} = 0,5.80 = 40(gam)\)

Bài 2 : 

 \(Zn + H_2SO_4 \to ZnSO_4 + H_2\)

Theo PTHH : 

\(n_{H_2SO_4} = n_{H_2} = n_{Zn} =\dfrac{13}{65} = 0,2(mol)\)

Suy ra :

\(V_{H_2} = 0,2.22,4 = 4,48(lít)\\ m_{H_2SO_4} = 0,2.98 = 19,6(gam)\)

n_Na2CO3 = 0,1 mol

Na₂CO₃ + CaCl₂ → CaCO₃ + 2NaCl

0,1..............0,1................0,1............0,2

⇒ m_CaCO3 = 0,1.100 = 10 (g)

⇒ m_NaCl = 0,2.58,5 = 11,7 (g)

8tk

a)

\(n_{H_2} = \dfrac{1,12}{22,4} = 0,05(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\)

Ta thấy : 

\(n_{HCl} = 0,2 > 2n_{H_2} = 0,1\) nên HCl dư.

Theo PTHH : \(n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{1}{30}(mol)\\ \Rightarrow m_{Al} = \dfrac{1}{30}.27 = 0,9(gam)\)

Ta có :

\(n_{HCl\ pư} = 2n_{H_2} = 0,1(mol)\\ \Rightarrow n_{HCl\ dư} = 0,2 - 0,1 = 0,1(mol)\\ \Rightarrow m_{HCl\ dư} = 0,1.36,5 = 3,65(gam)\)

Cho mình hỏi, ChO là chất gì thế ạ?

\(m_{tăng}=m_{O_2}=2.4\left(g\right)\)

\(n_{O_2}=\dfrac{2.4}{32}=0.075\left(mol\right)\)

\(V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)

\(Cu+\dfrac{1}{2}O_2\underrightarrow{t^0}CuO\)

\(0.15....0.075\)

\(m_{Cu}=0.15\cdot64=9.6\left(g\right)\)

Ta có :

n_SO2 = 4,48 : 22,4 = 0,2 (mol)

n_NaOH = 16 : 40 = 0,4 (mol)

Xét T = \(\dfrac{n_{NaOH}}{n_{SO2}}=\dfrac{0,4}{0,2}=2\)

=> Ta có PTHH :

2NaOH + SO₂ -----> Na₂SO₃ + H₂O

Ta có :

n_Na2SO3 = n_SO2 = 0,2 (mol)

=> m_Na2SO3 = 0,2 . (23 . 2 + 32 + 48) = 25,2(g)

a) PTHH : Zn + 2 HCl -> ZnCl2 + H2 

nH2= 0,15(mol)

-> nZn= nZnCl2=nH2= 0,15(mol)

-> Số nguyên tử kẽm p.ứ: 6.10²³ .0,15= 9.10²² (nguyên tử) 

b) mZnCl2= 136.0,15= 20,4(g)

c) H2+ 1/2 O2 -to-> H2O 

nH2= 0,15/2= 0,075(mol)

-> nH2O= nH2= 0,075(mol)

-> mH2O= 0,075.18= 1,35(g)

\(V_{hh}=V_{C_2H_2}+V_{H_2}=3V_{H_2}+V_{H_2}=4V_{H_2}=20\left(m^3\right)\)

\(\Rightarrow V_{H_2}=5\left(m^3\right),V_{C_2H_2}=15\left(m^3\right)\)

\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^0}2CO_2+H_2O\)

\(H_2+\dfrac{1}{2}O_2\underrightarrow{t^0}H_2O\)

\(V_{O_2}=\dfrac{5}{2}V_{C_2H_2}+\dfrac{1}{2}V_{H_2}=\dfrac{5}{2}\cdot15+\dfrac{1}{2}\cdot5=40\left(m^3\right)\)

Ta có :

\(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ n_{O_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)

Ta thấy : \(\dfrac{n_{Al}}{4} = 0,025 < \dfrac{n_{O_2}}{3} =0,03\) nên O₂ dư.

\(n_{Al_2O_3} = 0,5n_{Al} = 0,05(mol)\\ \Rightarrow m_{Al_2O_3} = 0,05.102 = 5,1(gam)\)

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

\(n_{O_2\left(đktc\right)}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: \(4Al+3O_2\rightarrow2Al_2O_3\)

Ta có tỉ lệ: \(\dfrac{0,1}{4}< \dfrac{0,1}{3}\)

-> \(O_2\) sẽ dư sau phản ứng.

Theo pthh: \(n_{Al_2O_3}=\dfrac{2}{4}n_{Al}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

-> \(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)