Bài 22: Tính theo phương trình hóa học

Gọi : CTHH là : \(H_xS_yO_z\)

\(x:y:z=\dfrac{\%H}{1}:\dfrac{\%S}{32}:\dfrac{\%O}{16}=\dfrac{2.04}{1}:\dfrac{32.65}{32}:\dfrac{65.31}{16}=2.04:1.02:4.08=2:1:4\)

CTHH là : H₂SO₄

Bài 1:

M_A = 42.2 = 84 (g/mol)

\(m_C=\dfrac{84.85,72}{100}=72\left(g\right)=>n_C=\dfrac{72}{12}=6\left(mol\right)\)

\(m_H=84-72=12\left(g\right)=>n_H=\dfrac{12}{1}=12\left(mol\right)\)

=> A là C₆H₁₂

\(n_{C_6H_{12}}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: \(C_6H_{12}+6O_2\underrightarrow{t^o}6CO_2+6H_2O\)

______0,3----->1,8_______________________(mol)

=> \(V_{O_2}=1,8.22,4=40,32\left(l\right)\)

Bài 2

\(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 2KClO₃ --to--> 2KCl + 3O₂

________0,4<-------------------------0,6___________(mol)

=> \(m_{KClO_3}=0,4.122,5=49\left(g\right)\)

a) \(n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\); \(n_{Al_2O_3}=\dfrac{61,2}{102}=0,6\left(mol\right)\)

PTHH: Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

Xét tỉ lệ: \(\dfrac{0,6}{1}>\dfrac{0,6}{3}\) => H₂SO₄ hết, Al₂O₃ dư

b) PTHH: Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

_______________0,6---------->0,2______________(mol)

=> \(m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\)

\(a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ b,n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\ \Rightarrow n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)\\ \Rightarrow V_{H_2}=0,3.22,4=6,72(l)\\ c,n_{AlCl_3}=n_{Al}=0,2(mol)\\ \Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)\)

\(a.CH_4+3O_2-^{t^o}\rightarrow CO_2+2H_2O\\ b.m_{CH_4}+m_{O_2}=m_{CO_2}+m_{H_2O}\\ c.a=m_{CO_2}+m_{H_2O}-m_{O_2}=4,4+3,6-6,4=1,6\left(g\right)\)

\(a,PTHH:4P+5O_2\xrightarrow{t^o}2P_2O_5\\ b,BTKL:m_{P}+m_{O_2}=m_{P_2O_5}\\ c,a=m_{O_2}=m_{P_2O_5}-m_{P}=14,2-6,2=8(g)\)

giúp mình câu 4 ạ. Cảm ơn nhiều ạ.