\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

a.

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,2------------------------>0,2

\(3H_2+Fe_2O_3\underrightarrow{t^o}2Fe+3H_2O\)

0,2------------>  \(\dfrac{2}{15}\)

b) \(V_{H_2\left(đkc\right)}=0,2.24,79=4,958\left(l\right)\)

c) 

Lập tỉ lệ: \(\dfrac{0,2}{3}>\dfrac{0,1}{1}\Rightarrow Fe_2O_3dư\)

\(m_{Fe}=\dfrac{2}{15}.56=7,47\left(g\right)\)

\(n_{H_2SO_4}=0,1.0,1=0,01\left(mol\right)\)

\(Na_2O+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

0,1<-----0,1

\(a=m_{Na_2O}=0,1.62=6,2\left(g\right)\)

đổi 1 tấn = 1000 kg

Pt CaCO₃ -> CaO + CO₂ 

Số mol CaCO₃ : 1000 : 100 = 1 mol 

=> nCaCO₃ = nCaO = nCO₂ 

Khối lượng CaO \(m=n.M.90\%=\dfrac{252}{5}\left(kg\right)\)

Thể tích CO₂ : \(V=n.22,4.90\%=20,16\left(l\right)\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{HCl}=0,5.2,1=1,05\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Zn}+3n_{Al}\Rightarrow n_{Zn}=0,225\left(mol\right)\)

⇒ m_Zn = 0,225.65 = 14,625 (g)

\(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,525\left(mol\right)\Rightarrow V_{H_2}=0,525.24,79=13,01475\left(l\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,225\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,225}{0,5}=0,45\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\end{matrix}\right.\)

a, Gọi: n_Fe2O3 = n_CuO = a (mol)

⇒ 160a + 80a = 24 ⇒ a = 0,1 (mol)

PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Theo PT: \(n_{H_2SO_4}=n_{CuO}+3n_{Fe_2O_3}=0,4\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{CuSO_4}=n_{CuO}=0,1\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ m _muối = m_CuSO4 + m_Fe2(SO4)3 = 0,1.160 + 0,1.400 = 56 (g)

c, \(C_{M_{CuSO_4}}=C_{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)

a, CTHH: O₂

M_O2 = 16.2 = 32 (g/mol)

b, CTHH: CaCO₃

M_CaCO3 = 40 + 12 + 16.3 = 100 (g/mol)

c, CTHH: Fe₂O₃

M_Fe2O3 = 56.2 + 16.3 = 160 (g/mol)

d, CTHH: Al(OH)₃

M_Al(OH)3 = 27 + (16 + 1).3 = 78 (g/mol)

16

Gọi CTHH cần tìm là N_xO_y

Ta có: M_NxOy = 54.2 = 108 (g/mol)

\(\Rightarrow x=\dfrac{108.25,93\%}{14}=2\)

Mà: 14x + 16y = 108

⇒ y = 5

Vậy: CTHH cần tìm là N_xO₅.

a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(n_{HCl}=\dfrac{200.3,65\%}{36,5}=0,2\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\), ta được Fe dư.

Theo PT: \(n_{FeCl_2}=n_{Fe\left(pư\right)}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)

b, m _{dd sau pư} = 11,2 - 0,1.56 + 200 - 0,1.2 = 205,4 (g)

\(\Rightarrow C\%_{FeCl_2}=\dfrac{0,1.127}{205,4}.100\%\approx6,2\%\)