1a) \(n_{CO_2}=0,14\left(mol\right);n_{Ca\left(OH\right)_2}=0,08\left(mol\right)\Rightarrow n_{OH^-}=0,16\left(mol\right)\)

\(\dfrac{n_{OH^-}}{n_{CO_2}}=\dfrac{0,16}{0,14}=1,14\) => Tạo 2 muối

\(Ca\left(OH\right)_2+2CO_2\rightarrow Ca\left(HCO_3\right)_2\)

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

b) Gọi x, y là số mol Ca(HCO3)2 và CaCO3

\(\left\{{}\begin{matrix}x+y=0,08\\2x+y=0,14\end{matrix}\right.\)

=> x=0,06 ; y=0,02

\(m_{CaCO_3}=0,02.100=2\left(g\right)\)

c) \(CM_{Ca\left(HCO_3\right)_2}=\dfrac{0,06}{0,8}=0,075M\)

2. a) Bảo toàn nguyên tố C : \(n_{CO_2}=n_{Na_2CO_3}=0,25\left(mol\right)\)

\(n_{KOH}=0,5\left(mol\right)\)

Lập T = \(\dfrac{n_{KOH}}{n_{CO_2}}=\dfrac{0,5}{0,25}=2\) => Tạo 1 muối K2CO3, các chất phản ứng hết

b) \(m_{K_2CO_3}=0,25.138=34,5\left(g\right)\)

\(n_{CO2}=\frac{3,136}{22,4}=0,14\left(mol\right)\)

\(n_{Ca\left(OH\right)2}=0,1.0,8=0,08\left(mol\right)\)

\(T=\frac{0,14}{0,08}=1,75\)

Vì 1 < T < 2 nên sản phẩm thu được gồm :CaCO3 , Ca(HCO3)2

PTHH

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

x_________________x_____________

\(2CO_2+Ca\left(OH\right)_2\rightarrow Ca\left(HCO_3\right)_2\)

2y______________________y

Gọi nCaCO3 = x và nCa(HCO3)2 = y

Theo PT ta có :
\(\left\{{}\begin{matrix}x+2y=0,14\\x+y=0,08\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,06\end{matrix}\right.\)

\(\Rightarrow m_{CaCO3}=0,02.100=2\left(g\right)\)

\(CM_{Ca\left(HCO3\right)2}=\frac{0,06}{0,8}=0,075M\)

Bài 1 : 

$n_{CO_2} = \dfrac{3,136}{22,4} = 0,14(mol)$
$n_{Ca(OH)_2} = 0,8.0,1 = 0,08(mol)$

CO₂ + Ca(OH)₂ → CaCO₃ + H₂O

0,08.......0,08...........0,08........................(mol)

CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂

0,06........0,06........................................(mol)

Suy ra : $m_{CaCO_3} = (0,08 - 0,06).100 = 2(gam)$

Bài 2 : 

$n_{CO_2} = \dfrac{2,24}{22,4} = 0,1(mol) ; n_{NaOH} = 0,1.1,5 = 0,15(mol)$
2NaOH + CO₂ → Na₂CO₃ + H₂O

0,15........0,075.......0,075....................(mol)

Na₂CO₃ + CO₂ + H₂O → 2NaHCO₃

0,025........0,025...................0,05..............(mol)

Suy ra:  

$C_{M_{NaHCO_3}} = \dfrac{0,05}{0,1} = 0,5M$
$C_{M_{Na_2CO_3}} = \dfrac{0,075 - 0,025}{0,1} = 0,5M$

b)

$NaOH + HCl \to NaCl + H_2O$
$n_{HCl} = n_{NaOH} = 0,15(mol)$
$m_{dd\ HCl} = \dfrac{0,15.36,5}{25\%} = 21,9(gam)$

\(n_{CO2}=0,14\left(mol\right),n_{Ca\left(OH\right)2}=0,08\left(mol\right)\)

\(1< \frac{n_{OH^-}}{n_{CO2}}< 2\rightarrow\) Tạo CaCO3; Ca(HCO3)2

Dặt nCaCO3=x; nCa(HCO3)2=y

\(\left\{{}\begin{matrix}2x+y=0,14\\x+y=0,08\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x=0,06\\y=0,02\end{matrix}\right.\)

\(m_{kt}=0,06.100=6\left(g\right)\)

Trong dd có Ca(HCO3)2

\(\rightarrow CM_{Ca\left(HCO3\right)2}=\frac{0,02}{0,8}=0,025M\)

a) \(n_{CO_2}=\dfrac{3,136}{22,4}=0,14\left(mol\right)\)

800ml = 0,8 lít

Có: C_M = \(\dfrac{n}{V}=\dfrac{n}{0,8}=0,1\) M

\(\Rightarrow n_{Ca\left(OH\right)_2}\) = 0,1 . 0,8 = 0,08 (mol)

CO₂ + Ca(OH)₂ -----> CaCO₃ + H₂O

1 1

0,14 0,08

Ta thấy: 0,14 > 0,08

=> Ca(OH)₂ hết, CO₂ dư

=> \(n_{CaCO_3}\) = 0,08 mol

=> \(m_{CaCO_3}\) = 0,08. 100 = 8g

b) C_{M dd sau phản ứng} = \(\dfrac{n}{V}=\dfrac{0,08}{0,8}=0,1M\)

n CO2 = \(\frac{3,136}{22,4}\)= 0,14 mol

nCa(OH)₂=0,1 x0,8 = 0,08 mol

Do \(\frac{1}{1}\)< \(\frac{nCo2}{nCa\left(OH\right)2}\)= \(\frac{7}{4}\)<\(\frac{2}{1}\)

nên có hai muối được tạo thành là CaCO3 và Ca(HCO3)₂. đặt số mol của CaCO₃ là a ( mol ); số mol của Ca(HCO₃)₂ là b ( mol )

CO₂ + Ca(OH)₂ ➜ CaCO3 + H₂O (1)

1 1 1 1 ( mol )

a a a a ( mol )

2CO₂ + Ca(OH)₂ ➜ Ca(HCO₃)₂ (2)

2 1 1 ( mol )

2b b b ( mol )

ta có hpt

\(\left\{{}\begin{matrix}a+2b=0,14\\a+b=0,08\end{matrix}\right.\)⇔b= 0,06; a= 0,02

theo pt (1) ta có m_kt = m_CaCO3 = a x M_CaCO3 = 0,02 x 100 = 2g

vậy m_kt = 2g

\(a.n_{CO_2}=\dfrac{0,672}{22,4}=0,03mol\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

\(n_{CO_2}=n_{Ca\left(OH\right)_2}=n_{CaCO_3}=0,03mol\\ m_{CaCO_3}=0,03.100=3g\\ b.V_{ddCa\left(OH\right)_2}=\dfrac{0,03}{1,5}=0,02l\)

n Al=0,25 mol

2Al + 6HCl → 2AlCl₃ + 3H₂

0,25---0,75------0,25------0,375 mol

=>VH2=0,375.22,4=8,4l

=>m AlCl3=0,25.133,5=33,375g

=>CM HCl=\(\dfrac{0,75}{2}\)=0,375M

\(n_{AgNO_3}=0,1.0,2=0,02\left(mol\right)\\ n_{HCl}=0,3.0,2=0,06\left(mol\right)\\ AgNO_3+HCl\rightarrow AgCl\downarrow\left(trắng\right)+HNO_3\\ a,Vì:\dfrac{0,02}{1}< \dfrac{0,06}{1}\Rightarrow HCldư\\ \Rightarrow n_{AgCl}=n_{HNO_3}=n_{AgNO_3}=0,02\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{AgCl}=143,5.0,02=2,87\left(g\right)\\ b,dd.sau.p.ứ:HNO_3,HCl\left(dư\right)\\ n_{HCl\left(dư\right)}=0,06-0,02=0,04\left(mol\right)\\V_{ddsau}=V_{ddAgNO_3}+V_{ddHCl}=0,2+0,2=0,4\left(l\right)\\ \Rightarrow C_{MddHCl\left(dư\right)}=\dfrac{0,04}{0,4}=0,1\left(M\right)\\ C_{MddHNO_3}=\dfrac{0,02}{0,4}=0,05\left(M\right)\)