a. \(n_{H_2}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

PTHH : Fe₂O₃ + 3H₂ -> 2Fe + 3H₂O

           0,5          1,5         1

\(m_{Fe_2O_3}=0,5.160=80\left(g\right)\)

b. \(m_{Fe}=1.56=56\left(g\right)\)

nH2 = 9.10^23/6.10^23 = 1,5 (mol)

PTHH: Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O

Mol: 0,5 <--- 1,5 ---> 1

mFe2O3 = 0,5 . 160 = 80 (g)

mFe = 1 . 56 = 56 (g)

Fe₂O₃+3H₂-to>2Fe+3H₂O

0,5-------1,5---------1 mol

n _H2=\(\dfrac{9,10^{23}}{6.10^{23}}\)=1,5 mol

=>m _Fe3O4=0,5.232=116g

=>m _Fe=1.56=56g

a, SO₃, Fe₂O₃, KNO₃ , BaSO₄

b, SO₃, NO₂, CO

4FeS+7O₂-to>2Fe₂O₃+4SO₂

  0,2----0,35------0,1 mol

n _Fe2O3=\(\dfrac{16}{160}\)=0,1 mol

=>V_O2=0,35.22,4=7,84l

=>m _FeS2=0,2.120=24g

\(1) Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O \text{Theo PTHH }\\ n_{H_2O} = n_{H_2} = \dfrac{20,16}{22,4}=0,9(mol)\\ \text{Bảo toàn khối lượng : }\\ a = m_{hh} + m_{H_2} - m_{H_2O} = 65,4 + 0,9.2 - 0,9.18 = 51(gam)\)

2)

\(n_{Mg} = a ; n_{Al} = b ; n_{Fe} = c\\ \Rightarrow 24a + 27b + 56c = 18,6(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = a + 1,5b + c = \dfrac{14,56}{22,4}=0,65(2)\\ 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ n_{O_2} = \dfrac{7,84}{22,4} = 0,35\)

Ta có :

\(\dfrac{a + b + c}{0,5a + 0,75b + \dfrac{2}{3}c} = \dfrac{0,55}{0,35}(3)\\ (1)(2)(3) \Rightarrow a = 0,2 ; b = 0,2 ; c= 0,15\\ \%m_{Mg} = \dfrac{0,2.24}{18,6}.100\% = 25,81\%\\ \%m_{Al} = \dfrac{0,2.27}{18,6}.100\% = 29,03\%\\ \%m_{Fe} = 100\% - 25,81\% -29,03\% = 45,16\%\)

a) \(n_{H_{2}SO_{4}}\)= \(\dfrac{0,3.10^{23}}{6.10^{23}}\)=0,05mol.

\(m_{H_{2}SO_{4}}\)= 0,05.98=4,9g

b)\(V_{CO_{2}}\)= 0,25.22,4=5,6l

Câu 5:

PTHH : H2+ Cl2 -to-> 2 HCl

Vì số mol , tỉ lệ thuận theo thể tích , nên ta có:

25/1 = 25/1 => P.ứ hết, không có chất dư, tính theo chất nào cũng được

=> V(HCl)= 2. V(H2)= 2. 25= 50(l)

Câu 4: mFe2O3= 0,6. 80= 48(g)

=> nFe2O3= 48/160=0,3(mol)

mCuO= 80-48=32(g) => nCuO=32/80=0,4(mol)

PTHH: CuO + CO -to-> Cu + CO2
0,4_______0,4_____0,4____0,4(mol)

Fe2O3 + 3 CO -to-> 2 Fe +3 CO2

0,3_____0,9____0,6______0,9(mol)

=>nCO= 0,4+ 0,9= 1,3(mol)

=> V(CO, đktc)= 1,3. 22,4=29,12(l)

\(n_X=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(n_{H_2O}=\dfrac{19.8}{18}=1.1\left(mol\right)\)

\(n_{CO_2}=3n_X=3\cdot0.4=1.2\left(mol\right)\)

\(m_{CO_2}=1.2\cdot44=52.8\left(g\right)\)

\(\text{Bảo toàn O : }\)

\(n_{O_2}=n_{CO_2}+\dfrac{1}{2}n_{H_2O}=1.2+\dfrac{1}{2}\cdot1.1=1.75\left(mol\right)\)

\(V_{O_2}=1.75\cdot22.4=39.2\left(l\right)\)

Bài 2 câu a sửa CaP thành CaO nhé, ghi nhầm..

a) n N2=\(\frac{0.54.10^{23}}{6.10^{23}}=0,09.10^{23}\)

b) n CO2=\(\frac{4,4}{44}=0,1\left(mol\right)\)

V CO2=0,1.22,4=2,24(l)

c) n N2O5=\(\frac{8,961}{22,4}=0,4\left(mol\right)\)

m N2O5=0,4.108=43,2(g)

d) m Fe(OH)3=1,5.107=160,5(g)