nP=0.2(mol)

nO2=0.3(mol)

4P+5O2->2P2O5

0.2  0.3

Ta có tỉ lệ: 0.2:4<0.3:5

->O2 dư tính theo P

nP2O5=0.1(mol)

->m=14.2(g)

Bạn xem lại đề giùm mình,9.6g O2 hay 9.6 l(đktc)

a)

\(n_{P_2O_5} = \dfrac{42,6}{142} = 0,3(mol)\\ \)

4P      +      5O₂ \(\xrightarrow{t^o}\)         2P₂O₅

0,6.............0,75.................0,3..........(mol)

m_P = 0,6.31 = 18,6(gam)

b)

2KClO₃  \(\xrightarrow{t^o}\)   2KCl     +    3O₂

0,5....................................0,75.....(mol)

\(m_{KClO_3} = 0,5.122,5 = 61,25(gam)\)

c)

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

\(n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)\\ \dfrac{n_{Fe}}{3} = 0,1 < \dfrac{n_{O_2}}{2} = 0,375\)

nên hiệu suất tính theo số mol Fe.

\(n_{Fe\ pư} = 0,3.90\% = 0,27(mol)\\ n_{Fe_3O_4} =\dfrac{1}{3}n_{Fe\ pư} = 0,09(mol)\\ \Rightarrow m_{Fe_3O_4} = 0,09.232 = 20,88(gam)\)

Câu 1: 

\(4P+5O_2\rightarrow2P_2O_5\)

\(n_P=\dfrac{15.5}{31}=0.5\left(mol\right)\)

\(\Leftrightarrow n_{P_2O_5}=0.25\left(mol\right)\)

\(\Leftrightarrow m_{P_2O_5}=0.25\cdot142=35.5\left(g\right)\)

Câu 1: 

4P + 5O₂ \(\underrightarrow{t^o}\) 2P₂O₅

\(n_P=\dfrac{15,5}{31}=0,5mol\)

\(n_{O_2}=\dfrac{0,5.5}{4}=0,625mol\)

\(V_{O_2}=0,625.22,4=14l\)

\(n_{P_2O_5}=\dfrac{0,5.2}{4}=0,25mol\\ m_{P_2O_5}=0,25.142=35,5g\)

Câu 2:

4P + 5O₂ \(\underrightarrow{t^o}\) 2P₂O₅

\(n_P=\dfrac{15,5}{31}=0,625mol\\ n_{P_2O_5}=\dfrac{0,625.2}{4}=0,25mol\\ m_{P_2O_5}=0,25.142=35,5g\)

\(Bài.2.có.nhiều.cách.làm.nhé.bạn\)

\(n_{O_2\left(đktc\right)}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 4P+5O_2\underrightarrow{^{to}}2P_2O_5\\ 0,12........0,15.........0,06\left(mol\right)\\ m_P=0,12.31=3,72\left(g\right)\)

Ta có: \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 4P + 5O₂ ---t^o---> 2P₂O₅.

Theo PT: n_P = \(\dfrac{4}{5}.n_{O_2}=\dfrac{4}{5}.0,15=0,12\left(mol\right)\)

=> m_P = 31 . 0,12 = 3,72(g)

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

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

Ta có: \(n_P=\dfrac{9,3}{31}=0,3\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,375\left(mol\right)\\n_{P_2O_5}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,375\cdot22,4=8,4\left(g\right)\\m_{P_2O_5}=0,15\cdot142=21,3\left(g\right)\end{matrix}\right.\)

a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

\(4P+5O_2--t^o->2P_2O_5\)

Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)

b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(CH_4+2O_2--t^o->CO_2+2H_2O\)

Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)

nO2=0,3mol

pthh: S+O2=>SO2

           0,3<-0,3->0,3

=> m=0,3.32=9,6g

V=0,3.22,4=6,72l

\(n_{Al_2O_3}=\dfrac{20.4}{102}=0.2\left(mol\right)\)

\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)

\(0.4.........0.3..........0.2\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)

mAl=0.4⋅27=10.8(g)mAl=0.4⋅27=10.8(g)

VO2=0.3⋅22.4=6.72(l)

vì iem đang dùng điện thoại nên ko gõ đc latex