a, \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right);n_{KOH}=0,1.2=0,2\left(mol\right)\)

PTHH: CO₂ + 2KOH → K₂CO₃ + 2H₂O

Mol:                   0,2          0,1

Ta có: \(\dfrac{0,15}{1}>\dfrac{0,2}{2}\) ⇒ CO₂ dư, KOH hết

Trong (F) chứa chất tan là K₂CO₃

b, \(m_{K_2CO_3}=0,1.138=13,8\left(g\right)\)

c, 

PTHH: K₂CO₃ ---t^o→ K₂O + CO₂

Mol:       0,1                      0,1

\(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

a, \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)

\(m_{NaOH}=0,2.40=8\left(g\right)\)

b, \(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)

\(c,C\%=\dfrac{6}{200}.100\%=3\%\)

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

Ta có: \(n_{NaOH}=0,1\cdot1,5=0,15\left(mol\right)\) \(\Rightarrow m_{NaOH}=0,15\cdot40=6\left(g\right)\)

Đáp án D

Gọi số mol HCl là x mol

HCl + KOH → KCl + H₂O

x                         x                           (mol)

Giả sử KOH hết ⇒ m_KCl = 74,5 . 0,1 = 7.45(g) > 6,525 ⇒KOH dư ,HCl hết.

(0,1 - x).56 + x.(39 + 35,5) = 6,525

⇒ x = 0,05 mol ⇒ C_M = 0,5M

m(Zn,Mg)=25-6,5= 18,5(g)

nHCl(p.ứ)= 0,8.2 : 125%= 1,28(mol)

PTHH: Zn + 2 HCl -> ZnCl2 + H2

x__________2x_____x____x(mol)

Mg +  2 HCl -> MgCl2 + H2

y______2y____y_____y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}65x+24y=18,5\\2x+2y=1,28\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{157}{2050}\\y=\dfrac{231}{410}\end{matrix}\right.\)

=> 

\(\%mAg=\dfrac{6,5}{25}.100=26\%\\ \%mZn=\dfrac{\dfrac{157}{2050}.65}{25}.100\approx19,912\%\\ \rightarrow\%mMg\approx54,088\%\)

Bài 14 : 

\(a) n_{CuO} = \dfrac{8}{80} = 0,1(mol)\\ CuO + 2HCl \to CuCl_2 + H_2O\\ n_{HCl} = 2n_{CuO} = 0,2(mol)\\ m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)\\ b) \text{Chất tan : } CuCl_2\\ n_{CuCl_2} = n_{CuO} = 0,1(mol)\\ m_{CuCl_2} = 0,1.135 = 13,5(gam)\)

Bài 15 : 

\(a) n_{Fe_2O_3} =\dfrac{4,8}{160} = 0,03(mol)\\ Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O\\ n_{H_2SO_4} = 3n_{Fe_2O_3} = 0,09(mol)\\ m_{dd\ H_2SO_4} = \dfrac{0,09.98}{9,8\%} = 90(gam)\\ b) \text{Chất tan : } Fe_2(SO_4)_3\\ n_{Fe_2(SO_4)_3} = n_{Fe_2O_3} = 0,03(mol)\\ m_{Fe_2(SO_4)_3} = 0,03.400 = 12(gam)\)

Bài 16 :

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\\ b) n_{HCl} = 2n_{Fe} = 0,4(mol)\\ C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%\)

a) MgSO4+2KOH---->Mg(OH)2+K2SO4

Mg(OH)2--->MgO+H2O

b)n\(_{MgSO4}=0,1.2=0,2\left(mol\right)\)

n\(_{KOH}=\frac{28}{56}=0,5\left(mol\right)\)

=> KOH dư

Theo pthh1

n\(_{Mg\left(OH\right)2}=n_{MgSO4}=0,2\left(mol\right)\)

m\(_{Mg\left(OH\right)2}=0,2.58=11,6\left(g\right)\)

c) Theo pthh

n\(_{K2SO4}=n_{MgSO4}=0,2\left(mol\right)\)

m\(_{K2SO4}=0,2.174=34,8\left(g\right)\)

Đổi: 100ml = 0,1 lít

\(\Rightarrow nMgSO_4=0,1.2=0,2mol\)

\(\Rightarrow nKOH=\frac{28}{56}=0,5mol\)

a, PTHH: \(MgSO_4+2KOH\rightarrow Mg\left(OH\right)_2+K_2SO_4\)

TBR: 1 : 2 : 1 : 1

TPT: 0,2 mol : 0,5 mol

0,2 mol \(\rightarrow\) 0,4 mol : 0,2 mol : 0,2 mol

Lập tỉ lệ: \(\frac{0,2}{1}< \frac{0,5}{2}\left(0,2< 0,25\right)\)

\(\Rightarrow\) KOH dư, tính toán dựa vào \(MgSO_4\)

b, \(\rightarrow\) \(mK_2SO_4=0,2.174=34,8g\)