Ta có: X gồm \(\left\{{}\begin{matrix}HCO_3^-:0,1a\left(mol\right)\\CO_3^{2-}:0,1\left(mol\right)\end{matrix}\right.\) và Y gồm: \(\left\{{}\begin{matrix}H^+:0,2\left(mol\right)\\SO_4^{2-}:0,025\left(mol\right)\\Cl^-:0,15\left(mol\right)\end{matrix}\right.\)

- Nhỏ X vào Y:

\(HCO_3^-+H^+\rightarrow CO_2+H_2O\)

k.0,1a___k.0,1a___k.0,1a (mol)

\(CO_3^{2-}+2H^+\rightarrow CO_2+H_2O\)

k.0,1____2.k.0,1____k.0,1 (mol)

n_H⁺ = k.0,1a + 2k.0,1 = 0,2

n_CO2 = k.0,1a + k.0,1 = 0,12

\(\Rightarrow\dfrac{k.0,1a+2k.0,1}{k.0,1a+0,1k}=\dfrac{0,2}{0,12}\)

⇒ a = 0,5 (M)

- Nhỏ Y và X:

\(CO_3^{2-}+H^+\rightarrow HCO_3^-\)

0,1______0,1______0,1 (mol)

\(HCO_3^-+H^+\rightarrow CO_2+H_2O\)

__0,15___0,1_____0,1 (mol)

→ Z gồm: SO₄^2-: 0,025 (mol) và HCO₃^-: 0,05 (mol)

- Cho Ba(OH)₂ dư vào:

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

_______0,025______0,025 (mol)

\(OH^-+HCO_3^-\rightarrow CO_3^{2-}+H_2O\)

_________0,05_____0,05 (mol)

\(Ba^{2+}+CO_3^{2-}\rightarrow BaCO_3\)

________0,05_______0,05 (mol)

⇒ m_↓ = m_BaSO4 + m_BaCO3 = 15,675 (g) = m

PT: \(CuO+C\underrightarrow{t^o}Cu+CO\)

\(Fe_2O_3+3C\underrightarrow{t^o}2Fe+3CO\)

\(2Al_2O_3+9C\underrightarrow{t^o}Al_4C_3+6CO\)

\(CaO+3C\underrightarrow{t^o}CaC_2+CO\)

\(Al_4C_3+12H_2O\rightarrow4Al\left(OH\right)_3+3CH_4\)

\(CaC_2+2H_2O\rightarrow Ca\left(OH\right)_2+C_2H_2\)

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

\(Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O\)

\(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

\(2Cu+O_2\underrightarrow{t^o}CuO\)

Ta có: \(n_{CuO}=n_{Cu}=n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

⇒ m_CuO = 0,2.80 = 16 (g)

\(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=\dfrac{1}{2}n_{H_2}=0,2\left(mol\right)\)

⇒ m_Fe2O3 = 0,2.160 = 32 (g)

Ta có: \(\left\{{}\begin{matrix}n_{CH_4}+n_{C_2H_2}=\dfrac{9,916}{24,79}=0,4\\16n_{CH_4}+26n_{C_2H_2}=4,625.4.0,4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,3\left(mol\right)\\n_{C_2H_2}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2O_3}=2n_{Al_4C_3}=2.\dfrac{1}{3}n_{CH_4}=0,2\left(mol\right)\\n_{CaO}=n_{CaC_2}=n_{C_2H_2}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ m_Al2O3 = 0,2.102 = 20,4 (g)

m_CaO = 0,1.56 = 5,6 (g)

Ta có: \(C_{N_1}=\dfrac{1000m}{E.V}=\dfrac{1000m}{E.1000}=0,01\)

\(C_{N_2}=\dfrac{1000m}{E.V}=\dfrac{1000m}{E.V}=0,2\)

\(\Rightarrow\dfrac{C_{N_1}}{C_{N_2}}=\dfrac{0,01}{0,2}=\dfrac{1}{20}\)

⇒ V = 50 (ml)

Đáp án: D

\(n_{HCl}=0,3.2=0,6\left(mol\right)\\ m_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,2}{1}< \dfrac{0,6}{2}\Rightarrow HCldư\\ n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2\left(đkc\right)}=24,79.0,2=4,958\left(l\right)\)

\(n_{HCl}=\dfrac{10\%.109,5}{36,5}=0,3\left(mol\right);n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,3}{2}>\dfrac{0,1}{1}\Rightarrow HCldư\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.0,1=0,2\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,3-0,2=0,1\left(mol\right)\\ V_{H_2\left(đkc\right)}=24,79.0,1=2,479\left(l\right)\\ b,ddA:HCl\left(dư\right),MgCl_2\\ m_{ddA}=2,4+109,5-0,1.2=111,7\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,1.36,5}{111,7}.100\%\approx3,268\%;C\%_{ddMgCl_2}=\dfrac{0,1.95}{111,7}.100\%\approx8,505\%\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right);n_{HCl}=\dfrac{365.10\%}{36,5}=1\left(mol\right)\\PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{1}{2}>\dfrac{0,1}{1}\Rightarrow HCldư\\ n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)

Câu 2:

Ta có: 80n_CuO + 160n_Fe2O3 = 16 (1)

m _giảm = 16.25% = 4 (g) = m_{O (trong oxit)}

\(\Rightarrow n_{O\left(trongoxit\right)}=\dfrac{4}{16}=0,25\left(mol\right)\)

BTNT O, có: n_CuO + 3n_Fe2O3 = 0,25 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CuO}=0,1\left(mol\right)\\n_{Fe_2O_3}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,1.80}{16}.100\%=50\%\\\%m_{Fe_2O_3}=50\%\end{matrix}\right.\)

Bạn bổ sung đủ đề câu 3 nhé.

Câu 1:

Ta có: \(n_{CO}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

BTNT C, có: n_CO2 = n_CO = 0,25 (mol)

BTKL, có: m_hh + m_CO = m _{chất rắn} + m_CO2

⇒ m _{chất rắn} = 30 + 0,25.28 - 0,25.44 = 26 (g)

Ta có: \(n_{NaOH}=\dfrac{40.10\%}{40}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{100.14,6\%}{36,5}=0,4\left(mol\right)\)

a, PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{1}\), ta được HCl dư.

Theo PT: \(n_{HCl\left(pư\right)}=n_{NaCl}=n_{NaOH}=0,1\left(mol\right)\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,1=0,3\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\)

b, \(C\%_{HCl}=\dfrac{10,95}{40+100}.100\%\approx7,82\%\)

\(C\%_{NaCl}=\dfrac{0,1.58,5}{40+100}.100\%\approx4,18\%\)

\(m_{CaO\left(lt\right)}=\dfrac{94,08}{80\%}\cdot100\%=117,6kg\\ CaCO_3\xrightarrow[]{t^0}CaO+CO_2\\ \Rightarrow\dfrac{m_{CaCO_3}}{100}=\dfrac{117,6}{56}\\ \Rightarrow m_{CaCO_3}=210kg\\ \%m_{CaCO_3\left(trong.đá.vôi\right)}=\dfrac{210}{280}\cdot100\%=75\%\)

PT: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

Ta có: \(n_{CaO}=\dfrac{94,08}{56}=1,68\left(kmol\right)\)

Theo PT: \(n_{CaCO_3\left(LT\right)}=n_{CaO}=1,68\left(kmol\right)\)

Mà: H = 80%
\(\Rightarrow n_{CaCO_3\left(TT\right)}=\dfrac{1,68}{80\%}=2,1\left(mol\right)\)

\(\Rightarrow m_{CaCO_3\left(TT\right)}=2,1.100=210\left(kg\right)\)

\(\Rightarrow\%CaCO_3=\dfrac{210}{280}.100\%=75\%\)