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

\(nO_2=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow nP_2O_5=\dfrac{2}{5}nO_2=\dfrac{2}{5}.0,1=0,04\left(mol\right)\)

\(\Rightarrow mP_2O_5=0,04.\left(31.2+16.5\right)=5,68\left(g\right)\)

c/

pthh: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\uparrow\)

\(nO_2=0,1\left(mol\right)\)

\(\Rightarrow nKMnO_4=\dfrac{2}{1}.nO_2=\dfrac{2}{1}.0,1=0,2\left(mol\right)\)

\(\Rightarrow mKMnO_4=0,2.\left(39+55+16.4\right)=31,6\left(g\right)\)

lỗi

lỗi

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

\(nO_2=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

\(\Rightarrow nP_2O_5=\dfrac{2}{5}.nO_2=\dfrac{2}{5}.0,1=0,04\left(mol\right)\)

\(mP_2O_5=0,04.142=5,68\left(g\right)\)

nO2 = 2,479/24,79 = 0,1 (mol)

PTHH: 4P + 5O2 -> (t°) 2P2O5

Mol: 0,08 <--- 0,1 ---> 0,04

mP2O5 = 0,05 . 142 = 5,68 (g)

nO2 = 2,479/24,79 = 0,1 (mol)

PTHH: 4P + 5O2 -> (t°) 2P2O5

Mol: 0,08 <--- 0,1 ---> 0,04

mP2O5 = 0,05 . 142 = 5,68 (g)

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

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

Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,2\left(mol\right)\Rightarrow m_{P_2O_5}=0,2.142=28,4\left(g\right)\)

c, \(n_{O_2}=\dfrac{5}{4}n_P=0,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)

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

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{3}\left(mol\right)\Rightarrow m_{KClO_3}=\dfrac{1}{3}.122,5=\dfrac{245}{6}\left(g\right)\)

n_O2 = \(\dfrac{1,12}{22,4}=0,05\) mol

Pt: 4P + 5O₂ --to--> 2P₂O₅

.............0,05 mol--> 0,02 mol

m_P2O5 thu được = 0,02 . 142 = 2,84 (g)

Pt: 2KMnO₄ --to--> K₂MnO₄ + MnO₂ + O₂

....0,1 mol<-----------------------------------0,05 mol

m_KMnO4 cần dùng = 0,1 . 158 = 15,8 (g)

n_O2 = 0,05 mol

4P + 5O₂ → 2P₂O₅

⇒ n_P2O5 = 0,025 mol

⇒ m_P2O5 = 0,025.142 = 3,55 (g)

2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

⇒ n_{KMnO4 =} 0,1 mol

⇒ m_KMnO4 = 0,1.158 = 15,8(g)

pt: 4P+5O2--->2P2O5

nO2=1,12/22,4=0,05(mol)

Theo pt: nP2O5=2/5nO2=2/5.0,05=0,02(mol)

=>mP2O5=0,02.142=2,84(g)

pt:2KMnO4--->KMnO4+MnO2+O2

Theo pt: nKMnO4=2nO2=2.0,05=0,1(mol)

=>mKMnO4=0,1.158=15,8(g)

a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

b, Ta có: \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

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

Có: O₂ hao hụt 40% → H% = 100 - 40 = 60%

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,4\left(mol\right)\)

\(\Rightarrow n_{KMnO_4\left(TT\right)}=\dfrac{0,4}{60\%}=\dfrac{2}{3}\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=\dfrac{2}{3}.158\approx105,3\left(g\right)\)

\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\\ a,PTHH:4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ b,n_{O_2}=\dfrac{5}{4}.0,4=0,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ c,n_{P_2O_5}=\dfrac{2}{4}.0,4=0,2\left(mol\right)\\ m_{P_2O_5}=142.0,2=28,4\left(g\right)\)

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

b, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,4}{5}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)

\(\Rightarrow m_{O_2\left(dư\right)}=0,15.32=4,8\left(g\right)\)

c, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

d, \(m_{P_2O_5}=14,2.80\%=11,36\left(g\right)\)

Giúp mik vs T-T