giúp mình với, mình sắp thi rồi

\(n_{Na_2CO_3}=\dfrac{360.21,2\%}{100\%.106}=0,72(mol)\\ n_{H_2SO_4}=2,5.0,2=0,5(mol)\\ PTHH:Na_2CO_3+H_2SO_4\to Na_2SO_4+H_2O+CO_2\uparrow\\ a,\text {Vì }\dfrac{n_{Na_2CO_3}}{1}>\dfrac{n_{H_2SO_4}}{1} \text {nên }Na_2CO_3\text { dư}\\ \Rightarrow n_{CO_2}=n_{H_2SO_4}=0,5(mol)\\ \Rightarrow V_{CO_2}=0,5.22,4=11,2(l)\\\)

\(b,A:Na_2SO_4\\ n_{Na_2SO_4}=n_{H_2SO_4}=0,5(mol)\\ m_{dd_{H_2SO_4}}=200.1,1=220(g);V_{dd_{Na_2CO_3}}=\dfrac{360}{1,2}=300(ml)=0,3(l)\\ \Rightarrow C\%_{Na_2SO_4}=\dfrac{0,5.142}{360+200-0,5.44}.100\%=13,2\%\\ C_{M_{Na_2SO_4}}=\dfrac{0,5}{0,3+0,2}=1M\)

\(n_{NaOH}=0,2.1=0,2\left(mol\right)\\ n_{H_2SO_4}=0,3.1,5=0,45\left(mol\right)\)

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

0,2------->0,1--------->0,1

Xét \(\dfrac{0,2}{2}< \dfrac{0,45}{1}\Rightarrow\) \(H_2SO_4\)dư

Trong dung dịch D có:

\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,45-0,1=0,35\left(mol\right)\\n_{Na_2SO_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,35}{0,5}=0,7M\\CM_{Na_2SO_4}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)

b

\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

0,35<---------0,35

\(V_{Ca\left(OH\right)_2}=\dfrac{0,35.74}{1,2}=\dfrac{259}{12}\approx21,58\left(ml\right)\\ \Rightarrow V_{dd.Ca\left(OH\right)_2}=\dfrac{\dfrac{259}{12}.100\%}{10\%}=\dfrac{1295}{6}\approx215,83\left(ml\right)\)

\(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(m_{ddNaOH}=1,3.500=650\left(g\right)\)

\(\dfrac{m_{ctNaOH}.100\%}{650}=25\%\Rightarrow m_{ctNaOH}=162,5\left(g\right)\)

\(n_{NaOH}=\dfrac{162,5}{40}=4,0625\left(mol\right)\)=>NaOH dư

\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

0,5-----------------------0,5---------------

\(C_M=\dfrac{0,5}{0,5}=1M\)

\(n_{H_2}=\dfrac{1,85925}{24,79}=0,075\left(mol\right)\)

\(m_{dd.HCl}=500.1,2=600\left(g\right)\)

\(2R+6HCl\rightarrow2RCl_3+3H_2\)

0,05<-0,15<--0,05<----0,075

a. \(R=\dfrac{1,35}{0,05}=27\left(g/mol\right)\)

Vậy tên kim loại là nhôm (Al)

b. 

\(n_{AgCl}=\dfrac{14,35}{143,5}=0,1\left(mol\right)\)

\(HCl+AgNO_3\rightarrow AgCl+HNO_3\)

0,025<-------------0,025

\(AlCl_3+3Ag\left(NO_3\right)\rightarrow3AgCl+Al\left(NO_3\right)_3\)

0,025------------------->0,075

\(CM_{HCl.đã.dùng}=\dfrac{0,025}{0,5}=0,05M\)

c.

\(m_{dd.X}=1,35+600-0,075.2=601,2\left(g\right)\)

\(n_{HCl.dư}=0,025.2=0,05\left(mol\right)\)

\(C\%_{AlCl_3}=\dfrac{0,05.133,5.100\%}{601,2}=1,11\%\)

\(C\%_{HCl.dư}=\dfrac{0,05.36,5.100\%}{601,2}=0,3\%\)

Bài 1: 

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\n_{NaOH}=\dfrac{164\cdot1,22\cdot20\%}{40}=1,0004\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo muối trung hòa

PTHH: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

Vì NaOH dư nên tính theo CO₂ \(\Rightarrow\left\{{}\begin{matrix}n_{Na_2CO_3}=0,25\left(mol\right)\\n_{NaOH\left(dư\right)}=0,5004\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Na_2CO_3\left(rắn\right)}=0,25\cdot106=26,5\left(g\right)\\m_{NaOH\left(rắn\right)}=0,5004\cdot40=20,016\left(g\right)\end{matrix}\right.\)

*Các bài còn lại bạn làm theo gợi ý bên dưới 
PTHH: \(CO_2+NaOH\rightarrow NaHCO_3\)  (1)

            \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)  (2)

a) \(NaOH+HCl\rightarrow NaCl+H_2O\)

0,1................0,3

LẬp tỉ lệ : \(\dfrac{0,1}{1}< \dfrac{0,3}{1}\)=> Sau pứ HCl dư

\(m_{NaCl}=0,1.58,5=5,85\left(g\right)\)

b) \(CM_{NaCl}=\dfrac{0,1}{0,2+0,3}=0,2M\)

\(CM_{HCl\left(dư\right)}=\dfrac{\left(0,3-0,1\right)}{0,2+0,3}=0,4M\)

Đây là.môn hoá nhé

Cho 5,6 lít CO2 (đktc) đi qua 164 ml dung dịch NaOH 20% (d = 1,22 g/ml) thu được dung dịch X. Cô cạn dung dịch X thì thu?