nO2=48/32=1,5(mol)

a) PTHH: C+ O2 -to-> CO2

nC=nO2=1,5(mol)

=> mC=nC.M(C)=1,5.12=18(g)

b) PTHH: S+ O2 -to-> SO2

nS=nO2=1,5(mol)

=> mS=nS.M(S)=1,5.32= 48(g)

c) PTHH: 4P + 5 O2 -to->2  P2O5

nP= 4/5. nO2=4/5 . 1,5=1,2(mol)

=>mP=1,2.31=37,2(g)

Chúc em học tốt!

\(n_{O_2}=\dfrac{48}{32}=1.5\left(mol\right)\)

\(a.\)

\(C+O_2\underrightarrow{^{^{t^0}}}CO_2\)

\(n_C=n_{O_2}=1.5\left(mol\right)\)

\(m_C=1.5\cdot12=18\left(g\right)\)

\(b.\)

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

\(n_S=n_{O_2}=1.5\left(mol\right)\)

\(m_S=1.5\cdot32=48\left(g\right)\)

\(c.\)

\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)

\(n_P=\dfrac{4}{5}\cdot n_{O_2}=\dfrac{4}{5}\cdot1.5=1.2\left(mol\right)\)

\(m_P=1.2\cdot31=37.2\left(g\right)\)

\(a) n_P = \dfrac{18,6}{31} = 0,6(mol)\\ n_{O_2} = \dfrac{20,16}{22,4} = 0,9(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ \dfrac{n_P}{4} = 0,15 < \dfrac{n_{O_2}}{5} = 0,18 \to O_2\ dư\\ n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,75(mol)\\ \Rightarrow m_{O_2\ dư} = (0,9-0,75).32 = 4,8(gam)\\ b) n_{Fe} = \dfrac{56}{56} = 1(mol)\)

\(3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ \dfrac{n_{Fe}}{3} = \dfrac{1}{3}<\dfrac{n_{O_2}}{2} = 0,45\to Fe\ cháy\ hết.\\ c)\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,9.2 = 1,8(mol)\\ \Rightarrow m_{KMnO_4} = 1,8.158 =284,4(gam)\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\) ; \(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

0,2  <  0,25                        ( mol )

0,2      \(\dfrac{2}{15}\)                 \(\dfrac{1}{15}\)   ( mol )

`->` Chất dư là O₂

\(m_{O_2\left(dư\right)}=\left(0,25-\dfrac{2}{15}\right).32=3,73\left(g\right)\)

\(V_{kk}=V_{O_2}.5=\dfrac{2}{15}.22,4.5=14,93\left(l\right)\)

\(m_{bôt.sắt}=\dfrac{11,2.100}{100-12}=12,72\left(g\right)\)

ta tính nO2=44,8:22,4=2mol

          4<-2

=> cần 4 mol Cacbon

                    8/5<-2

vậy cần 1,6mol Photpho

b) Fe+O2=>FeO

    2<-2

=> cần 2 mol bột sắt

\(n_P=\dfrac{m}{M}=0,2\left(mol\right)\)

- Ta có : \(V_{O_2}=\dfrac{V_{kk}}{5}=4,48\left(l\right)\)

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

\(4P+5O_2\rightarrow2P_2O_5\)

- Theo phương pháp đường chéo ta có :

=> Sau phản ứng O₂ phản ứng hết, P còn dư ( dư 0,04 mol )

Vậy sau phản ứng photpho không cháy hết .

b, - Chất được tạo thành là P2O5 .

Theo PTHH : \(n_{P2O5}=\dfrac{n_P}{2}=\dfrac{0,16}{2}=0,08\left(mol\right)\)

\(\Rightarrow m_{P2O5}=n.M=11,36\left(g\right)\)

\(n_{H_2}=\dfrac{17.92}{22.4}=0.8\left(mol\right)\)

\(n_{Fe_3O_4}=\dfrac{69.6}{232}=0.3\left(mol\right)\)

\(Fe_3O_4+4H_2\underrightarrow{t^0}3Fe+4H_2O\)

\(0.2..............0.8\)

\(m_{Fe_3O_4\left(dư\right)}=\left(0.3-0.2\right)\cdot232=23.2\left(g\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.8......................................0.8\)

\(m_{Zn}=0.8\cdot65=52\left(g\right)\)

a)

\(n_{O_2} = \dfrac{11,2}{22,4} = 0,5(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ n_P = \dfrac{4}{5}n_{O_2} = 0,4(mol)\\ \Rightarrow m_P = 0,4.31 = 12,4(gam)\)

b)

\(n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,2(mol)\\ \Rightarrow m_{P_2O_5} = 0,2.142 = 28,4(gam)\)

c)

\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,5.2 = 1(mol)\\ \Rightarrow m_{KMnO_4} = 1.158 = 158(gam)\)

a) nKMnO4= 63,2/158=0,4(mol)

nFe=0,5(mol)

PTHH: KMnO4 -to-> K2MnO4 + MnO2 + 1/2 O2

0,4________________________________0,2(mol)

=>V(O2,đktc)=0,2.22,4=4,48(l)

PTHH: Fe + 2 HCl -> FeCl2 + H2

0,5_____________________0,5(mol)

=> V(H2,đktc)=0,5.22,4=11,2(l)

b) H2 + 1/2 O2 -to-> H2O

0,5/2 > 0,2/1

=> H2 dư, O2 hết, tính theo nO2

=> nH2(dư)=0,5 - 0,2.2=0,1(mol)

=>mH2(dư)=0,1.2=0,2(g)

c) nFe3O4=166/232= 83/116(mol)

PTHH: Fe3O4 + 4 H2 -to-> 3 Fe + 4 H2O

nFe3O4(p.ứ)=0,5/4=0,125(mol)

=> mFe3O4=0,125.232=29(g)

nFe=3/4. 0,5=0,375(mol)

=>mFe=0,375.56=21(g)

      n_S=m_S/M_S=3,2/32=0,1(mol)
      n_O2=V_O2/22,4=32/22,4=1,42(mol)
PTHH: S + O₂ --> SO₂ (1)
BĐ:    0,1   1,42
PỨ:   0,1-->0,1-->0,1
SPỨ:  0--->0,32-->0,1
a) Từ PT(1)=>O₂ dư
_VO2(dư)=n_O2(dư) .22,4=0,32 .22,4=7,168(l)
b) Từ PT(1)=>n_SO2=0,1(mol)
=>m_SO2=n.M=0,1 .64=6,4(g)