\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ f,SO_2+H_2O\rightarrow H_2SO_3\\ SO_3+H_2O\rightarrow H_2SO_4\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\\ 2FeCl_3+3Ca\left(OH\right)_2\rightarrow2Fe\left(OH\right)_3+3CaCl_2\\ K_2SO_3+Cu\left(NO_3\right)_2\rightarrow2KNO_3+CuSO_3\)

e) CaO+…H2O. ---> Ca(OH)₂

f) SO2 + H2O --->  ...H2SO3

g)..SO3. + H20 ---> H2SO4

h) .. P2O5+ 3H2O ---> 2H3PO4

k) Sat (III) clorua + Canxi hiđroxit-> Sắt (III) hiđroxit + Canxi clorua

2FeCl3+3Ca(OH)2->2Fe(OH)3+3CaCl2

1) Kali sunfit+Đồng (II) nitrat =>Kali nitrat + Đồng (II) sunfit

K2SO3+Cu(NO3)2->2KNO3+CuSO3

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ SO_2+H_2O\rightarrow H_2SO_3\\ SO_3+H_2O\rightarrow H_2SO_4\\ P_2O_5+3H_2O\underrightarrow{t^o}2H_3PO_4\) 
\(2FeCl_3+3Ca\left(OH\right)_2\rightarrow2Fe\left(OH\right)_3+3CaCl_2\)
\(K_2SO_3+Cu\left(NO_3\right)_2\rightarrow2KNO_3+CuSO_3\)

a)

PTHH : \(SO_2+Ca\left(OH\right)_2\rightarrow CáO_4+H_2O\)

b)

Ta có :

\(n_{SO_2}=\frac{0,224}{22,4}=0,01\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0,01\times1,4=0,014\)

Theo ptpư : \(n_{SO_2}=n_{Ca\left(OH\right)_2}=n_{CaSO_3}=n_{H_2O}\)

Vậy n_{Ca(OH)2 ( dư )} = \(n_{Ca\left(OH\right)_2\left(bđ\right)}-n_{Ca\left(OH\right)_2\left(pư\right)}\)

\(=0,014-0,001=0,004\left(mol\right)\)

Pt : \(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)

Định luật bảo toàn khối lượng : 

\(m_{Ca}+m_{H2O}=m_{Ca\left(OH\right)2}+m_{H2}\)

    4    + \(m_{H2O}\) = 7,4 + 0,2

   4 + \(m_{H2O}\)    = 7,6

⇒ \(m_{H2O}\) = 7,6 - 4 = 3,6(g)

 Chúc bạn học tốt

\(a,PTHH:SO_2+Ca(OH)_2\to CaSO_3\downarrow+H_2O\\ b,n_{Ca(OH)_2}=0,7.0,01=0,007(mol)\\ n_{SO_2}=\dfrac{0,112}{22,4}=0,005(mol)\)

Vì \(\dfrac{n_{SO_2}}{1}<\dfrac{n_{Ca(OH)_2}}{1}\) nên \(Ca(OH)_2\) dư

\(\Rightarrow n_{CaSO_3}=n_{H_2O}=0,005(mol)\\ \Rightarrow m_{CaSO_3}=0,005.120=0,6(g)\\ m_{H_2O}=0,005.18=0,09(g)\)

a) PTHH:
So2 + Ca(OH)2 \Rightarrow CaSO3 (kt) + H2O
b) 
n SO2 =112/22400 =0.005 (mol)
n Ca(OH)2 =0.7 *0.01 =0.007 (mol)
Ca(OH)2 dư và dư : 0.007 -0.005 =0.002 (mol)
Theo phương trình hoá học :

SO2 + Ca(OH)2 \Rightarrow CaSO3 (kt) + H2O
1 mol 1 mol 1 mol
0.005 0.005 0.005
m CaCo3 =0.005 *120 =0.6 (g)
m Ca(OH)2 = 0.002 *74 =0.148 (g)

\(n_{SO_2}=\frac{\frac{112}{1000}}{22,4}==0,005\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=\frac{700}{1000}.0,01=0,007\left(mol\right)\)

a/ PTHH : Ca(OH)₂ + SO₂ -----> CaSO₃ + H₂O

 b/ Ta có : \(\frac{n_{Ca\left(OH\right)_2}\left(\text{đề bài}\right)}{n_{Ca\left(OH\right)_2}\left(\text{phương trình}\right)}=\frac{0,007}{1}>\frac{n_{SO_2}\left(\text{đề bài}\right)}{n_{SO_2}\left(\text{phương trình}\right)}=\frac{0,005}{1}\)

=> SO₂ phản ứng hết, Ca(OH)₂ dư 0,007 - 0,005 = 0,002 (mol)

+) \(n_{H_2O}=n_{SO_2}=0,005\left(mol\right)\Rightarrow m_{H_2O}=0,005.18=0,09\left(g\right)\)

+) \(n_{CaSO_3}=n_{SO_2}=0,005\left(mol\right)\Rightarrow m_{CaSO_3}=0,005.120=0,6\left(g\right)\)

+) \(m_{Ca\left(OH\right)_2\text{dư}}=0,002.74=0,148\left(g\right)\)

\(n_{SO_2}=\dfrac{0,112}{22,4}=0,005\left(mol\right)\\ n_{Ca\left(OH\right)_2}=0,01\cdot0,7=0,007\left(mol\right)\\ PTHH:SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3\downarrow+H_2O\\ \text{Vì }\dfrac{n_{SO_2}}{1}< \dfrac{n_{Ca\left(OH\right)_2}}{1}\text{ nên sau p/ứ }Ca\left(OH\right)_2\text{ dư}\\ \Rightarrow n_{CaSO_3}=n_{H_2O}=0,005\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{CaSO_3}=0,005\cdot120=0,6\left(g\right)\\m_{H_2O}=0,005\cdot18=0,09\left(g\right)\end{matrix}\right.\)

112ml = 0,112l 

700ml = 0,7 l

n SO2 = 0,112/ 22,4 =0,005 mol

n Ca(OH)2 = 0,7 . 0,01 = 0,007 mol

ta có tỉ lệ : n SO2 / n Ca(OH)2 = 0,005 / 0,007 =0,714 < 1 ---> pứ tạo muối CaSO3 

pthh SO2 + Ca(OH)2 ---> CaSO3 + H2O

theo pthh n CaSO3 = n Ca(OH)2 = 0,007 mol

---> m CaSO3 = 0,007 . 120= 0,84 g