Al2O3 + 6 HNO3 -> 2 Al(NO3)3 + 3 H2O 

a) nAl2O3= 10,2/102= 0,1(mol)

nAl(NO3)3= 0,1 x 2= 0,2(mol)

VddAl(NO3)3= VddHNO3= (0,1 x6)/ 1= 0,6(l)

CMddAl(NO3)3= 0,2/0,6= 0,333(M)

b) HNO3+ NaOH -> NaNO3 + H2O 

nNaOH= nHNO3= 0,6(mol)

-> mNaOH= 0,6 x 40= 24(g)

-> m=mddNaOH= 24/10% = 240(g)

-> m= 240(g)

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Theo PT:  \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)

a, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)

b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)

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

Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

a. PTHH: \(Zn+H_2SO_4--->ZnSO_4+H_2\uparrow\left(1\right)\)

b. Theo PT₍₁₎: \(n_{Zn}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=65.0,3=19,5\left(g\right)\)

c. Theo PT₍₁₎: \(n_{H_2SO_4}=n_{Zn}=0,3\left(mol\right)\)

Đổi 300ml = 0,3 lít

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,3}=1M\)

d. PTHH: \(2NaOH+H_2SO_4--->Na_2SO_4+2H_2O\left(2\right)\)

Theo PT₍₂₎: \(n_{NaOH}=2.n_{H_2SO_4}=2.0,3=0,6\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,6.40=24\left(g\right)\)

\(\Rightarrow m_{dd_{NaOH}}=\dfrac{24.100\%}{20\%}=120\left(g\right)\)

\(2HCl+Ba\left(OH\right)_2\rightarrow BaCl_2+2H_2O\)

Theo PT : \(n_{HCl}=2n_{Ba\left(OH\right)_2}=2.\dfrac{400.1,2.17,1\%}{171}=0,96\left(mol\right)\)

=> \(V_{HCl}=\dfrac{0,96.36,5}{3,65\%.1,05}=914,29\left(ml\right)\)

\(m_{Ba\left(OH\right)_2}=400\cdot1.2\cdot17.1\%=82.08\left(g\right)\)

\(n_{Ba\left(OH\right)_2}=\dfrac{82.08}{171}=0.48\left(mol\right)\)

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

\(0.48..............0.96\)

\(m_{HCl}=0.96\cdot36.5=35.04\left(g\right)\)

\(m_{dd_{HCl}}=\dfrac{35.04}{3.65\%}=960\left(g\right)\)

\(V_{dd_{HCl}}=\dfrac{960}{1.05}=1008\left(ml\right)\)

gọi số mol của NaOH = x , Ba(OH)2 = y . số mol OH- = x + 2y

pt : H+ + OH- → H2O 

nH+ =2. 0,035. 2=0,14 mol = x+ 2y 

khối lượng kết tủa = m_SO42-  + m_Ba2+ = 96.0.07 + 137.y= 9,32 =>y=0,02 => x= 0,1

nồng độ tự tính.

B1:

2NaOH+H₂SO₄\(\rightarrow\)Na₂SO₄+2H₂O

n_NaOH=\(\frac{4}{40}=0.1\)mol

=>n_H2SO4=\(\frac{1}{2}\)n_NaOH=0.05 mol

=>C_M=\(\frac{n_{H2SO42}}{V}\)=\(\frac{0.05}{200}\)=2,5.10^-4 (M)

B2:

Mg+\(\frac{1}{2}\)O₂\(\underrightarrow{t^0}\)MgO                (1)

MgO+2HCl\(\rightarrow\)MgCl₂+H₂O (2)

n_Mg(1)=\(\frac{0,36}{24}=0,015mol\)

=>n_MgO(1)=0,015=n_MgO(2)

n_HCl(2)=2n_MgO(2)=0,03mol

=>C_M(HCl)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)

https://hoc24.vn/cau-hoi/cho-400-g-dung-dich-naoh-30-tac-dung-vua-het-voi-100-g-dung-dich-hcl-tinh-a-nong-do-muoi-thu-duoc-sau-phan-ungb-tinh-nong-do-axit-hcl-bie.7974814552205

Tui trả lời rùi nghen

Theo giả thiết ta có : nCO2 = 6,72/22,4 = 0,3 (mol)

a) PTHH :

0,3mol......0,3mol................0,3mol.........0,3mol

b) nồng độ mol của dd Ba(OH)2 đã dùng là :

Câu 15: 

Bảo toàn nguyên tố: \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)=n_{BaCO_3}=n_{Ba\left(OH\right)_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{BaCO_3}=0,3\cdot197=59,1\left(g\right)\\C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,3}{0,6}=0,5\left(M\right)\end{matrix}\right.\)