Question

Answer 1

Bài 4

a) \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

______0,2-->0,4----->0,2---->0,2

V = 0,2.22,4 = 4,48(l)

b) m_FeCl2 = 127.0,2 = 25,4(g)

m_{HCl dư} = (0,5-0,4).36,5 = 3,65 (g)

Bài 6: 

a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right);n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

______0,4->0,3----------->0,2

=> X chứa Al₂O₃: 0,2 mol

=>m_X = 0,2.102 = 20,4(g)

b) n_{O2 dư} = 0,4 - 0,3 = 0,1 (mol)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

_____0,08<-0,1

=> m_P = 0,08.31 = 2,48(g)

Answer 2

Bài 4:

\(n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ a,Fe+2HCl\to FeCl_2+H_2\\ LTL:\dfrac{0,2}{1}<\dfrac{0,5}{2}\Rightarrow HCl\text{ dư}\\ \Rightarrow n_{H_2}=n_{FeCl_2}=0,2(mol);n_{HCl(dư)}=0,5-0,4=0,1(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\\ b,m_{FeCl_2}=0,2.127=25,4(g)\\ m_{HCl(dư)}=0,1.36,5=3,65(g)\)

Bài 5:

\(n_{Al}=\dfrac{10,8}{27}=0,4(mol);n_{O_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ 4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ LTL:\dfrac{0,4}{4}<\dfrac{0,4}{3}\Rightarrow O_2\text{ dư}\\ a,X:Al_2O_3,O_2\\ n_{Al_2O_3}=0,2(mol);n_{O_2(dư)}=0,4-0,3=0,1(mol)\\ \Rightarrow m_X=0,2.102+0,1.32=23,6(g)\\ b,4P+5O_2\xrightarrow{t^o}2P_2O_5\\ \Rightarrow n_P=0,08(mol)\\ \Rightarrow m_P=0,08.31=2,48(g)\)