\(n_{NaOH}=0,1.2=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ a.Muối:Na_2SO_4\\ b.n_{H_2SO_4}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m_{ddH_2SO_4}=\dfrac{0,1.98.100}{9,8}=100\left(g\right)=0,1\left(kg\right)\)

PT: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

Ta có: \(n_{H_2SO_4}=0,25.1=0,25\left(mol\right)\)

a, Theo PT: \(n_{NaOH}=2n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow V_{NaOH}=\dfrac{0,5}{2}=0,25\left(l\right)\)

b, Theo PT: \(n_{Na_2SO_4}=n_{H_2SO_4}=0,25\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,25}{0,25+0,25}=0,5\left(M\right)\)

Cu(OH)₂ + H₂SO₄ \(\rightarrow\) CuSO₄ + 2H₂O

nCu(OH)₂ = \(\dfrac{29,4}{98}=0,3mol\)

Theo pt: nH₂SO₄ = nCu(OH)₂ = 0,3 mol

=> mH₂SO₄ = 0,3.98 = 29,4g

VH₂SO₄ = 0,3:1 = 0,3l

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

Ta có: \(n_{CuO}=\dfrac{29,4}{80}=0,3675\left(mol\right)=n_{CuSO_4}=n_{H_2SO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuSO_4}=0,3675\cdot160=58,8\left(g\right)\\m_{H_2SO_4}=0,3675\cdot98=36,015\left(g\right)\\V_{H_2SO_4}=\dfrac{0,3675}{1}=0,3675\left(l\right)=367,5\left(ml\right)\end{matrix}\right.\)

nAl= 0,04(mol)

PTHH: 2 Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2

0,04___________0,06___0,02_____0,06(mol)

a) V(H2, đktc)=0,06.22,4=1,344(l)

b) VddH2SO4= 0,06/2=0,03(l)=30(ml)

c) VddAl2(SO4)3=VddH2SO4=0,03(l)

=>CMddAl2(SO4)3=0,02/0,03=2/3(M)

\(n_{Al}=\dfrac{1.08}{27}=0.04\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(0.04......0.06.............0.02...........0.06\)

\(V_{H_2}=0.06\cdot22.4=1.344\left(l\right)\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.06}{2}=0.03\left(l\right)\)

\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0.02}{0.03}=\dfrac{2}{3}\left(M\right)\)

a) 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

b) \(m_{NaOH}=\dfrac{200.8}{100}=16\left(g\right)\)

=> \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

                0,4--->0,2--------->0,2

=> \(m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)

c) \(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)

=> \(m_{dd.H_2SO_4}=\dfrac{19,6.100}{9,8}=200\left(g\right)\)

mNaOH = 8% . 200 = 16 (g)

nNaOH = 16/40 = 0,4 (mol)

PTHH: 2NaOH + H2SO4 -> Na2SO4 + 2H2O 

Mol: 0,4 ---> 0,2 ---> 0,2 ---> 0,4

mNa2SO4 = 0,2 . 119 = 23,8 (g)

mH2SO4 = 0,2 . 98 = 19,6 (g)

mddH2SO4 = 19,6/9,8% = 200 (g)

\(Mg + H_2SO_4 \rightarrow MgSO_4 + H_2\)

\(n_{Mg}= \dfrac{4,8}{24}= 0,2 mol\)

Theo PTHH:

\(n_{H_2}= n_{Mg} = 0,2 mol\)

\(\Rightarrow V_{H_2}= 0,2 . 22,4=4,48l\)

b)

Theo PTHH:

\(n_{H_2SO_4}= n_{Mg}= 0,2 mol\)

\(\Rightarrow V_{H_2SO_4}= \dfrac{0,2}{2}=0,1 l\)

\(a.\\ n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+H_2SO_4->MgSO_4+H_2\\ =>n_{H_2}=0,2\left(mol\right)\\ =>V=0,2\cdot22,4=4,48\left(l\right)\)

b.

Thể tích dung dịch \(H_2SO_4\) đã dùng là:

\(v_{H_2SO_4}=\dfrac{0,2}{2}=0,1\left(l\right)\)

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

a. PTHH: Mg + H₂SO₄ ---> MgSO₄ + H₂

Theo PT: \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)

=> \(V_{H_2}=0,2.22,4=4,48\left(lít\right)\)

b. Theo PT: \(n_{H_2SO_4}=n_{Mg}=0,2\left(mol\right)\)

Ta có: \(C_{M_{H_2SO_4}}=\dfrac{0,2}{V_{dd_{H_2SO_4}}}=2M\)

=> \(V_{dd_{H_2SO_4}}=0,1\left(lít\right)\)

A. Sai, do X là pentapeptit nên X chỉ có 4 liên kết peptit

B. Phần trăm khối lượng nitơ trong A là 20,29%. 

C. Sai, do A có 5 gốc amino axit trong phân tử

D. Sai
Chọn B