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

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

____0,2__0,25_____0,1 (mol)

⇒ V_O2 = 0,25.24,79 = 6,1975 (l)

m_P2O5 = 0,1.142 = 14,2 (g)

Ta có: \(n_{CH_4}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

\(n_{C_2H_4}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

\(CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\)

Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=0,7\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,7.24,79=17,353\left(l\right)\)

\(n_{BaCO_3}=n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=0,4\left(mol\right)\)

\(\Rightarrow m_{BaCO_3}=0,4.197=78,8\left(g\right)\)

Ta có: \(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

____0,02____0,04_________0,02 (mol)

⇒ m_HCl = 0,04.36,5 = 1,46 (g)

V_H2 = 0,02.24,79 = 0,4958 (l)

Bài 5:

Chất rắn không tan là Cu.

⇒ m_Cu = 1,86 (g) \(\Rightarrow\%m_{Cu}=\dfrac{1,86}{6}.100\%=31\%\)

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: 27n_Al + 56n_Fe = 6 - 1,86 (1)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{3,34665}{24,79}=0,135\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,06\left(mol\right)\\n_{Fe}=0,045\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,06.27}{6}.100\%=27\%\\\%m_{Fe}=100-31-27=42\%\end{matrix}\right.\)

Bạn bổ sung đề bài 3 nhé.

Bài 4:

a, Ta có: 160n_Fe2O3 + 80n_CuO = 24 (1)

PT: \(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)

\(CuO+CO\underrightarrow{t^o}Cu+CO_2\)

Theo PT: \(\left\{{}\begin{matrix}n_{Fe}=2n_{Fe_2O_3}\\n_{Cu}=n_{CuO}\end{matrix}\right.\) ⇒ 56.2n_Fe2O3 + 64n_CuO = 17,6 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=0,1\left(mol\right)\\n_{CuO}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ n_Fe = 0,1.2 = 0,2 (mol) ⇒ m_Fe = 0,2.56 = 11,2 (g)

m_Cu = 0,1.64 = 6,4 (g)

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

\(\Rightarrow V_{CO}=0,4.24,79=9,916\left(l\right)\)

c, - Phương pháp vật lý: dùng nam châm tách Fe ra khỏi hỗn hợp, thu được Cu.

- Phương pháp hóa học: Cho hỗn hợp kim loại phản ứng với dd HCl loãng, Cu không tan → đem sấy khô.

Ta có: \(n_{CO_2}=n_{CaCO_3}=\dfrac{59,1}{197}=0,3\left(mol\right)\)

m _{dd giảm} = m_CaCO3 - m_CO2 - m_H2O

⇒ 40,68 = 59,1 - 0,3.44 - 18n_H2O

⇒ n_H2O = 0,29 (mol)

BTNT C và H, có: \(\left\{{}\begin{matrix}6n_{C_6H_{12}O_6}+12n_{C_{12}H_{22}O_{11}}=0,3\\12n_{C_6H_{12}O_6}+22n_{C_{12}H_{22}O_{11}}=0,29.2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{C_6H_{12}O_6}=0,03\left(mol\right)\\n_{C_{12}H_{22}O_{11}}=0,01\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{C_{12}H_{22}O_{11}}=\dfrac{0,01.342}{0,01.342+0,03.180}.100\%\approx38,8\%\)

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, \(\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{5,85}{250}.100\%=2,34\%\\C\%_{BaCl_2}=\dfrac{10,4}{250}.100\%=4,16\%\end{matrix}\right.\)

b, \(C\%_{Fe\left(NO_3\right)_2}=\dfrac{0,3.180}{268}.100\%\approx20,15\%\)

c, \(C\%_{AlCl_3}=\dfrac{20}{20+100}.100\%\approx16,67\%\)

d, \(C\%_{K_3PO_4}=\dfrac{0,1.212}{0,1.212+200}.100\%\approx9,6\%\)

e, \(\left\{{}\begin{matrix}C\%_{KOH}=\dfrac{20}{200+300}.100\%=4\%\\C\%_{Ba\left(OH\right)_2}=\dfrac{30}{200+300}.100\%=6\%\end{matrix}\right.\)

f, \(\left\{{}\begin{matrix}C\%_{KOH}=\dfrac{0,2.56}{300+200}.100\%=2,24\%\\C\%_{Ba\left(OH\right)_2}=\dfrac{0,3.171}{300+200}.100\%=10,26\%\end{matrix}\right.\)

g, \(n_{HCl}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow C\%_{HCl}=\dfrac{0,1.36,5}{0,1.36,5+100}.100\%\approx3,52\%\)

h, \(m_{KOH}=100.10\%+200.20\%=50\left(g\right)\)

\(\Rightarrow C\%_{KOH}=\dfrac{50}{100+200}.100\%\approx16,67\%\)

i, \(m_{Ca\left(OH\right)_2}=50.20\%+100.10\%=20\left(g\right)\Rightarrow C\%_{Ca\left(OH\right)_2}=\dfrac{20}{50+100}.100\%\approx13,33\%\)

Ta có: m _oxit = m_KL + m_O

⇒ 7,8 = 5,8 + m_O

⇒ n_O = 0,125 (mol)

Pư với HCl: O + 2H → H₂O

⇒ n_HCl = 2n_O = 0,25 (mol)

\(\Rightarrow V_{HCl}=\dfrac{0,25}{1}=0,25\left(l\right)\)

Gọi: n_Al = 2a (mol), n_A = 3a (mol)

Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

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

\(A+2HCl\rightarrow ACl_2+H_2\)

Theo PT: n_HCl = 2n_H2 = 1,2 (mol)

\(\Rightarrow m_{ddHCl}=\dfrac{1,2.36,5}{21,9\%}=200\left(g\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_A\) ⇒ 0,6 = 3/2.2a + 3a ⇒ a = 0,1 

⇒ n_Al = 0,2 (mol), n_A = 0,3 (mol)

⇒ 0,2.27 + 0,3.M_A = 12,6 ⇒ M_A = 24 (g/mol)

→ A là Mg.

Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\end{matrix}\right.\)

Ta có: m _{dd sau pư} = 12,6 + 200 - 0,6.2 = 211,4 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,2.133,5}{211,4}.100\%\approx12,6\%\\C\%_{MgCl_2}=\dfrac{0,3.95}{211,4}.100\%\approx13,5\%\end{matrix}\right.\)