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

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

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

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

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)

\(n_{O_2}=\dfrac{V_{O_2}}{24,79}=\dfrac{6,21}{24,79}=0,25mol\)

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

0,2     0,25              0,1      ( mol )

\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,1.142=14,2g\)

\(m_P=n_P.M_P=0,2.31=6,2g\)

Câu 1

\(n_{Fe}=\dfrac{16,8}{56}=0,3mol\\ n_{O_2}=\dfrac{2,479}{24,79}=0,1mol\\ 3Fe+2O_2\xrightarrow[t^0]{}Fe_3O_4\\ \Rightarrow\dfrac{0,3}{3}>\dfrac{0,1}{2}\Rightarrow Fe.dư\\ 3Fe+2O_2\xrightarrow[t^0]{}Fe_3O_4\)

0,15       0,1         0,05

\(m_{Fe.dư}=16,8-0,15.56=8,4g\\ b.m_{Fe_3O_4}=0,05.232=11,6g\)

Câu 2

\(n_{H_2}=\dfrac{7,437}{24,79}=0,3mol\\ R+H_2SO_4\rightarrow RSO_4+H_2\\ n_{H_2}=n_R=0,3mol\\ M_R=\dfrac{12}{0,3}=40g/mol\)

Vậy M là Canxi

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(Cu+2H_2SO_{4\left(đ\right)}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

\(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(n_{SO_2}=\dfrac{2,9748}{24,79}=0,12\left(mol\right)\)

\(n_{Cu}=n_{SO_2}=0,12\left(mol\right)\)

\(\Rightarrow m=m_{Zn}+m_{Cu}=0,1.65+0,12.64=14,18\left(g\right)\)

Có: \(n_{H_2SO_{4\left(đ\right)}}=2n_{SO_2}=0,24\left(mol\right)\Rightarrow x=m_{ddH_2SO_4\left(đ\right)}=\dfrac{0,24.98}{98\%}=24\left(g\right)\)

n _C3H8=0,1 mol

C₃H₈+5O₂-to>3CO₂+4H₂O

0,1-----0,5------------0,3-----0,4 mol

=>V_CO2=0,3.22,4=6,72 l

=>V_O2=0,5.22,4=11,2l

a) nZn = 26/65 = 0,4 (mol)

PTHH: 2Zn + O2 -> (t°) 2ZnO

LTL: 0,4/2 < 0,3 => O2 dư

nO2 (p/ư) = 0,4/2 = 0,2 (mol)

mO2 (dư) = (0,3 - 0,2) . 32 = 3,2 (g(

b) nZnO = 0,4 (mol)

mZnO = 0,4 . 81 = 32,4 (g)

a. \(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)

\(n_{O_2}=\dfrac{7.437}{24,79}=0,3\left(mol\right)\)

Ta thấy : 0,4 > 0,3 => Zn dư , O₂ đủ

PTHH : 2Zn + O₂ ----t^o---> 2ZnO

            0,6    0,3                 0,6

\(m_{Zn\left(dư\right)}=\left(0,4-0,6\right).65=-13\left(g\right)\) 

b. \(m_{ZnO}=0,6.81=48,6\left(g\right)\)

\(n_{O_2}=\dfrac{2,479}{24,79}=0,1(mol)\\ 2Cu+O_2\xrightarrow{t^o}2CuO\\ \Rightarrow n_{CuO}=0,2(mol)\\ \Rightarrow m_{CuO}=0,2.80=16(g)\)

\(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\)

PTHH: 2KClO₃ -t^o, MnO₂-> 2KCl + 3O₂

             0,1--------------------->0,1--->0,15

=> \(\left\{{}\begin{matrix}V_{O_2}=0,15.24,79=3,7185\left(l\right)\\m_{KCl}=0,1.74,5=7,45\left(g\right)\end{matrix}\right.\)

\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

LTL: \(\dfrac{0,1}{4}< \dfrac{0,15}{5}\) => P có cháy hết

Ta có: