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a) PTHH         2Mg + O2 ===> 2MgO

b) Phương trình bảo toàn khối lượng là:

       m_MgO + m_O2 = m_MgO

c) Áp dụng định luật bào toàn khối lượng theo câu b) ta có:

   m_O2 = m_MgO - m_Mg

<=> m_O2 = 15 - 9 = 6 gam

\(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{19,5}{65}=0,3mol\)

\(Zn+\dfrac{1}{2}O_2\rightarrow\left(t^o\right)ZnO\)

1      1/2                   1       (mol)

0,3     0,15                   0,3        ( mol )

PƯ trên thuộc loại phản ứng hóa hợp

\(m_{ZnO}=n_{ZnO}.M_{ZnO}=0,3.81=24,3g\)

\(V_{O_2}=n_{O_2}.22,4=0,15.22,4=3,36l\)

a. \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

PTHH : 2Zn + O₂ -> 2ZnO   ( Phản ứng hóa hợp )

             0,3      0,15      0,3

b. \(m_{ZnO}=0,3.81=24,3\left(g\right)\)

c. \(V_{O_2}=0,15.22,4=3,36\left(l\right)\)

a) nFe= 0,25(mol)

PTHH: Fe + H2SO4 -> FeSO4 + H2

0,25______0,25______0,25__0,25(mol)

b) V(H2,đktc)=0,25.22,4=5,6(l)

c) mH2SO4= 0,25.98= 24,5(g)

Lời giải "chi tiết"

PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

Ta có: \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)=n_{H_2SO_4}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,25\cdot98=24,5\left(g\right)\\V_{H_2}=0,25\cdot22,4=5,6\left(l\right)\end{matrix}\right.\)

a,PTHH:

=0,25(mol)

Theo pt: nH2=nFe=0,25(mol)

⇒VH2=0,25.22,4=5,6(l)

c, Theo pt: nH2SO4=nFe=0,25(mol)

⇒mH2SO4=0,25.98=24,5(g)

a. Iron  tan dần, có khí không màu thoát ra

b. Iron + hydrochloric acid → iron (II) chloride + khí hydrogen 

c. 

Bảo toàn khối lượng : 

\(m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\)

\(\Rightarrow m_{HCl}=205+0.2-5.6=199.6\left(g\right)\)

d. 

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

a) 4Al + 3O₂ --t^o--> 2Al₂O₃

b) \(n_{O_2}=\dfrac{19,2}{32}=0,6\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

           0,8<-0,6---------->0,4

=> m_Al = 0,8.27 = 21,6(g)

c) m_Al2O3 = 0,4.102 = 40,8(g)

\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

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

0,2      0,15    0,1

\(m_{Al_2O_3}=0,1\cdot102=10,2g\)

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

0,1                            0,15

\(m_{KClO_3}=0,1\cdot122,5=12,25g\)

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

\(n_{Al}=5,4:27=0,2\left(mol\right)\)

\(\Rightarrow n_{Al_2O_3}=0,2.2:4=0,1\left(mol\right);n_{O_2}=0,2.3:4=0,15\left(mol\right)\)

\(m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

b)\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

\(n_{O_2}=0,15\left(mol\right)\)(câu a)

\(\Rightarrow n_{KClO_3}=0,15.2:3=0,1\left(mol\right)\)

\(m_{KClO_3}=0,1.123,5=12,35\left(g\right)\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Zn + 2HCl ---> FeCl₂ + H₂

          0,3<---------------0,3<----0,3

=> \(\left\{{}\begin{matrix}m=0,3.65=19,5\left(g\right)\\m_{muối}=0,3.136=40,8\left(g\right)\\V_{ddHCl}:thiếu.C_M\end{matrix}\right.\)

\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O
LTL: \(0,2>\dfrac{0,3}{3}\) => Fe₂O₃ dư

Theo pthh: n_{Fe2O3 (pư)} = \(\dfrac{1}{3}n_{H_2}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)

                  n_Fe = \(\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\)

=> m_{chất rắn} = 0,1.160 + 0,2.56 = 27,2 (g)

Zn+2HCl->ZnCl2+H2

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

n H2=0,3 mol

=>m Zn=0,3.65=19,5g

=>m muối=0,3.136=40,8g

c) thiếu đề

d)

Fe2O3+3H2-to>2Fe+3H2O

               0,3------0,2

n Fe2O3=0,2 mol

=>Fe2O3 dư

=>m cr=0,2.56+0,1.160=27,2g

\(a) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ n_{Al} = \dfrac{5,4}{27} = 0,2(mol)\\ n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,1(mol) \Rightarrow m_{Al_2O_3} = 0,1.102 = 10,2(gam)\\ b) n_{O_2} = \dfrac{3}{4}n_{Al} = 0,15(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,3(mol) \Rightarrow m_{KMnO_4} = 0,3.158 = 47,4(gam)\)

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

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

b, Theo PT:  \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\)

PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=0,2.158=47,4\left(g\right)\)

Bạn tham khảo nhé!

a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

b, \(n_P=\dfrac{1,55}{31}=0,05\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,05}{4}>\dfrac{0,05}{5}\), ta được P dư.

c, Theo PT: \(n_{P\left(pư\right)}=\dfrac{4}{5}n_{O_2}=0,04\left(mol\right)\Rightarrow n_{P\left(dư\right)}=0,05-0,04=0,01\left(mol\right)\)

\(\Rightarrow m_{P\left(dư\right)}=0,01.31=0,31\left(g\right)\)