a)

\(3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\)

b)

Ta có :

\(n_{Fe} = \dfrac{8,4}{56} = 0,15(mol)\\ n_{O_2} = \dfrac{96}{32} = 3(mol)\)

Ta thấy : \(\dfrac{n_{Fe}}{3} = 0,05 < \dfrac{n_{O_2}}{2} = 1,5\) do đó O₂ dư.

Theo PTHH :

\(n_{O_2\ pư} = \dfrac{2}{3}n_{Fe} = 0,1(mol)\\ \Rightarrow n_{O_2\ dư} = 3 - 0,1 = 2,9(mol)\\ \Rightarrow m_{O_2\ dư} = 92,8(gam)\)

c)

\(n_{Fe_3O_4} = \dfrac{1}{3}n_{Fe} = 0,05(mol)\\ \Rightarrow m_{Fe_3O_4} = 0,05.232 = 11,6(gam)\)

\(a)PTHH:FeCl_3+2O_2\xrightarrow[]{t^o}Fe_3O_4\)

mol                 1          2              1

mol

\(b)\)Số mol \(FeCl_3\) là: \(n_{FeCl_3}=\dfrac{m_{FeCl_3}}{M_{FeCl_3}}=\dfrac{8,4}{162,5}=0,052\left(mol\right)\)

Số mol \(O_2\) là: \(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{96}{32}=3\left(mol\right)\)

Lập tỉ lệ: \(\dfrac{1}{0,052}>\dfrac{2}{3}\Rightarrow FeCl_3dư\)

Số mol \(FeCl_3\) phản ứng là:

Từ PTHH\(\Rightarrow\) \(n_{FeCl_3}=\dfrac{0,052\times3}{3}=0,035\left(mol\right)\)

Số mol \(FeCl_3\) dư là: \(n_{FeCl_3dư}=n_{FeCl_3đầu}-n_{FeCl_3p/ứng}=0,052-0,035=0,018\left(mol\right)\)

Khối lượng \(FeCl_3\) dư là: \(m_{FeCl_3dư}=n_{FeCl_3dư}\times M_{FeCl_3}=0,018\times162,5=2,925\left(g\right)\)

a)

\(4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\)

b)

Ta thấy :

\(\dfrac{n_P}{4} = \dfrac{\dfrac{12,4}{31}}{4} =0,1 < \dfrac{n_{O_2}}{5} = \dfrac{\dfrac{20}{32}}{5} = 0,125\) 

do đó, O₂ dư

\(n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,5(mol)\\ \Rightarrow n_{O_2\ dư} = \dfrac{20}{32} - 0,5 = 0,125(mol)\)

c)

\(n_{P_2O_5} = \dfrac{1}{2}n_P = 0,2(mol)\\ \Rightarrow m_{P_2O_5} =0,2.142 = 28,4(gam)\)

a)

4P+5O2to→2P2O54P+5O2→to2P2O5

b)

Ta thấy :

nP4=12,4314=0,1<nO25=20325=0,125nP4=12,4314=0,1<nO25=20325=0,125 

do đó, O₂ dư

nO2 pư=54nP=0,5(mol)⇒nO2 dư=2032−0,5=0,125(mol)nO2 pư=54nP=0,5(mol)⇒nO2 dư=2032−0,5=0,125(mol)

c)

nP2O5=12nP=0,2(mol)⇒mP2O5=0,2.142=28,4(gam)

Bài 1.

a.\(n_{KClO_3}=\dfrac{49}{122,5}=0,4mol\)

\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

   0,4                                            0,6     ( mol )

\(V_{O_2}=0,6.22,4=13,44l\)

b.\(n_P=\dfrac{12,4}{31}=0,4mol\)

\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

\(\dfrac{0,4}{4}\)< \(\dfrac{0,6}{5}\)                          ( mol )

0,4                           0,2            ( mol )

Chất dư là O2

\(m_{O_2\left(dư\right)}=0,6-\left(\dfrac{0,4.5}{4}\right)=0,1mol\)

\(m_{P_2O_5}=0,2.142=28,4g\)

Bài 2.

a.\(n_{KMnO_4}=\dfrac{126,4}{158}=0,8mol\)

\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

    0,8                                                        0,4  ( mol )

\(V_{O_2}=0,4.22,4=8,96l\)

b.\(n_P=\dfrac{12,4}{31}=0,4mol\)

\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

\(\dfrac{0,4}{4}\) > \(\dfrac{0,4}{5}\)                      ( mol )

            0,4              0,16      ( mol )

Chất dư là P

\(n_{P\left(dư\right)}=0,4-\left(\dfrac{0,4.4}{5}\right)=0,08mol\)

\(m_{P_2O_5}=0,16.142=22,72g\)

\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)

\(n_{O_2}=\dfrac{0.448}{22.4}=0.02\left(mol\right)\)

\(4Na+O_2\underrightarrow{^{^{t^0}}}2Na_2O\)

\(4..........1\)

\(0.2.....0.02\)

\(LTL:\dfrac{0.2}{4}>\dfrac{0.02}{1}\Rightarrow Nadư\)

\(m_{Na\left(dư\right)}=\left(0.2-0.08\right)\cdot23=2.76\left(g\right)\)

\(m_{Na_2O}=0.04\cdot62=2.48\left(g\right)\)

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ pthh:S+O_2\underrightarrow{t^o}SO_2\\ LTL:\dfrac{0,1}{1}>\dfrac{0,05}{1}\) 
=> S dư 
\(n_{S\left(P\text{Ư}\right)}=n_{SO_2}=n_{O_2}=0,05\left(mol\right)\\ m_S=\left(0,1-0,05\right).32=1,6\left(g\right)\\ V_{SO_2}=0,05.22,4=1,12\left(l\right)\)

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

b, Ta có: \(n_{P_2O_5}=\dfrac{7,1}{142}=0,05\left(mol\right)\)

Theo PT: \(n_P=2n_{P_2O_5}=0,1\left(mol\right)\)

\(\Rightarrow m_P=0,1.31=3,1\left(g\right)\)

\(n_{O_2}=\dfrac{5}{2}n_{P_2O_5}=0,125\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)

c, Có: \(V_{O_2\left(dư\right)}=2,8.15\%=0,42\left(l\right)\)

\(\Rightarrow V_{O_2}=2,8+0,42=3,22\left(l\right)\)

Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

a. PTHH: 2Mg + O₂ ---> 2MgO

Ta thấy: \(\dfrac{0,1}{2}< \dfrac{0,5}{1}\)

Vậy oxi dư, magie hết.

Theo PT: \(n_{O_2}=\dfrac{1}{2}.n_{Mg}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

=> \(m_{O_{2_{dư}}}=0,5.32-0,05,32=14,4\left(g\right)\)

b. Theo PT: \(n_{MgO}=n_{Mg}=0,1\left(mol\right)\)

=> \(m_{MgO}=0,1.40=4\left(g\right)\)