a, PTPƯ: SO₃ + H₂O ---> H₂SO₄

n_SO3=\(\dfrac{2,24}{22,4}=0,1mol\)

1 mol SO₃ ---> 0,1 mol H₂SO₄

nên 0,1 mol SO₃ ---> 0,1 mol H₂SO₄

C_{M H2SO4}=\(\dfrac{0,1}{0,5}\)=0,2 M

b, PTPƯ: Zn + H₂SO₄ ---> ZnSO₄ + H₂

1 mol H₂SO₄ ---> 1 mol Zn

nên 0,1 mol H₂SO₄ ---> 0,1 mol Zn

m_Zn=0,1.65=6,5 g

\(n_{SO_3}=\dfrac{20}{80}=0,25\left(mol\right)\\ PTHH:SO_3+H_2O\rightarrow H_2SO_4\\ Mol:0,25\rightarrow0,25\rightarrow0,25\\ C_{MH_2SO_4}=\dfrac{0,25}{0,5}=0,5M\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\\ Mol:0,25\leftarrow0,25\\ m_{Mg}=0,25.24=6\left(g\right)\)

\(n_{Na}=\dfrac{6,9}{23}=0,3\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b,m_{ddsaup.ứ}=m_{Na}+m_{H_2O}-m_{H_2}=6,9+100-0,15.2=106,6\left(g\right)\)

\(n_P=\dfrac{6,2}{31}=0,2mol\)

\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

0,2     0,25                        0,1    ( mol )

\(V_{O_2}=0,25.22,4.\left(100+30\right)\%=7,28l\)

\(m_{H_2O}=\dfrac{235,8}{18}=13,1mol\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

 0,1   <   13,1                         ( mol )

0,1                            0,2            ( mol )

\(m_{ddspứ}=\left(0,1.142\right)+235,8=250g\)

\(C\%_{H_3PO_4}=\dfrac{0,2.98}{250}.100=7,84\%\)

\(V_{H_3PO_4}=\dfrac{0,2.98}{1,25}=15,68ml=0,01568l\)

\(C_M=\dfrac{0,2}{0,01568}=12,75M\)

n_H2SO4 = 49/98 = 0.5 (mol) 

C_MH2SO4 = 0.5/0.15 = 3.3 (M) 

Zn + H₂SO₄ => ZnSO₄  + H₂

...........0.5.............0.5.........0.5

V_H2 = 0.5 * 22.4 = 11.2 (l) 

C_MZnSO4 = 0.5 / 0.15 = 10/3 (M) 

C%_ZnSO4 = C_M*M / 10D = 10/3 * 161 / 10 * 1.25 = 42.9 %

\(n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ a,n_{NaOH}=n_{H_2O}2.0,05=0,1\left(mol\right)\\ \Rightarrow m_{NaOH}=0,1.40=4\left(g\right)\\ b,C1:n_{Na_2SO_4}=n_{H_2SO_4}=0,05\left(mol\right)\\ m_{sp}=m_{Na_2SO_4}+m_{H_2O}=0,05.142+18.0,1=8,9\left(g\right)\\ C2:m_{sp}=m_{H_2SO_4}+m_{NaOH}=4,9+4=8,9\left(g\right)\)

a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)

Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

\(n_{H_2O}=\dfrac{200}{18}=\dfrac{100}{9}\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{\dfrac{100}{9}}{2}\), ta được H₂O dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)

b, Theo PT: \(n_{NaOH}=n_{Na}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 2,3 + 100 - 0,05.2 = 102,2 (g)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,1.40}{102,2}.100\%\approx3,91\%\)

c, - Dung dịch làm quỳ tím hóa xanh.

\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{0,1}{2}=0,05\left(mol\right);n_{NaOH}=n_{Na}=0,1\left(mol\right)\\ a,V=V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ b,m_{ddNaOH}=m_{Na}+m_{H_2O}-m_{H_2}=2,3+200-0,05.2=202,2\left(g\right)\\ C\%_{ddNaOH}=\dfrac{40.0,1}{202,2}.100\approx1,978\%\\ c,NaOH-Tính.bazo\Rightarrow Quỳ.tím.hoá.xanh\)

Chọn đáp án D.

→ nOH^- trong X = 0,28 + 0,18×2 = 0,64

→ nCO₃^2- sinh ra = 0,64 - 0,45 = 0,19 > 0,18

→ nBaCO₃ = 0,18

→ mBaCO₃ = 0,18×197 = 35,46 gam.

Phương trình : 2K + 2H2O -> 2KOH + H2

                        0,4                      0,4      0,2

a) nH2 = 4,48 : 22,4 = 0,2 mol

b) nK = 0,4 mol

mK = 0,4 . 39 = 15,6

c) khối lượng sản phẩm thu đc sau PỨ là KOH

nKOH = 0,4 

mKOH = 0,4 . 56 = 22,4