Cho 5,6g CaO tác dụng với nước theo PTHH: CaO + H2O -> Ca(OH2). Tính khối lượng sản phẩm thu được.
PTHH: CaO + H2O -> Ca(OH2)
nCaO= 5,6/56= 0,1 (mol)
Theo ptpu: nCaO= nCa(OH)2= 0,1 (mol)
==>mCa(OH)2= 0,1.74= 7,4 (g)

theo bài ra ta có : m_CaO =5,6 g

=> n_CaO = 0,1mol

theo pt : n_Ca(OH)2 = 0,1 mol

=> m_CaO = 0,1 . 74 = 7,4 g

n_CaO = \(\dfrac{5,6}{56}\)= 0,1mol

CaO + H₂O -> CA(OH)₂

0,1 ->0,1

=>m_Ca(OH)2 = 0,1. 74 = 7,4 g

a) 

\(n_{CaO}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PTHH: CaO + H₂O --> Ca(OH)₂

             0,15----------->0,15

=> m_Ca(OH)2 = 0,15.74 = 11,1 (g)

b) \(C_M=\dfrac{0,15}{0,5}=0,3M\)

c) 

PTHH: 2Ca + O₂ --t^o--> 2CaO

                     0,075<----0,15

=> V_O2 = 0,075.24,79 = 1,85925 (l)

\(a,n_{CaO}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PTHH: CaO + H₂O ---> Ca(OH)₂ 

            0,15-------------->0,15

=> m_Ca(OH)2 = 0,15.74 = 11,1 (g)

b, \(C_{M\left(Ca\left(OH\right)_2\right)}=\dfrac{0,15}{0,5}=0,3M\)

c, PTHH: 2Ca + O₂ --t^o--> 2CaO

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

=> V_O2 = 0,075.24,79 = 1,85925 (l)

CaO + H2O  -- > Ca(OH)2

nCaO = 8,4 / 56 = 0,15 (mol)

mCa(OH)2  = 0,15 . 74 = 11,1 (g)

CM Ca(OH)2 = 0,15 / 0,5 = 0,3 M

2Ca + O2 -- > 2CaO

 nO2 = 1/2 . nCaO = 1/2 . 0,15 = 0,075 (mol)

VO2 = 24,79 . 0,075 = 1,85925 (l)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)

b, Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

c, \(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,25}{1}\), ta được CuO dư.

Theo PT: \(n_{CuO\left(pư\right)}=n_{H_2}=0,2\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,25-0,2=0,05\left(mol\right)\)

\(\Rightarrow m_{CuO\left(dư\right)}=0,05.80=4\left(g\right)\)

\(a) Ca + 2H_2O \to Ca(OH)_2 + H_2\\ CaO + H_2O \to Ca(OH)_2\\ n_{Ca} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ m_{Ca} = 0,15.40 = 6(gam)\\ \Rightarrow m_{CaO} = 17,2 - 6 = 11,2(gam)\\ b) n_{Ca(OH)_2} = n_{Ca} + n_{CaO} = 0,15 + \dfrac{11,2}{56} = 0,35(mol)\\ m_{Ca(OH)_2} = 0,35.74 = 25,9(gam)\)

\(a,n_{Na}=\dfrac{2,53}{23}=0,11\left(mol\right)\)

PTHH: 2Na + 2H₂O ---> 2NaOH + H₂

           0,11--------------->0,11

=> m_NaOH = 0,11.40 = 4,4 (g)

\(b,n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

PTHH: Na₂O + H₂O ---> 2NaOH

            0,25---------------->0,5

=> m_NaOH = 0,5.40 = 20 (g)

\(c,n_{Ca}=\dfrac{12}{40}=0,3\left(mol\right)\)

PTHH: Ca + 2H₂O ---> Ca(OH)₂ + H₂

           0,3---------------->0,3

=> m_Ca(OH)2 = 0,3.74 = 22,2 (g)

\(d,n_{CaO}=\dfrac{1,12}{56}=0,02\left(mol\right)\)

PTHH: CaO + H₂O ---> Ca(OH)₂

           0,02------------->0,02

=> m_Ca(OH)2 = 0,02.74 = 1,48 (g)

 c

C

a, \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

b, \(n_{CaO}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

\(n_{H_2O}=\dfrac{100}{18}=\dfrac{50}{9}\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{\dfrac{50}{9}}{1}\), ta được H₂O dư.

Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaO}=0,05\left(mol\right)\Rightarrow m_{Ca\left(OH\right)_2}=0,05.74=3,7\left(g\right)\)

Chọn A

Đáp án cần chọn là: A

5,6 hay 6,5 :) ?

a, nH2 = 5,6/22,4 = 0,25 (mol)

PTHH: 2H2 + O2 -> (t°) 2H2O

Mol: 0,25 ---> 0,125 ---> 0,25

VO2 = 0,125 . 22,4 = 2,8 (l)

b, mH2O = 0,25 . 18 = 4,5 (g)

c, PTHH: S + O2 -> (t°) SO2

nSO2 = nO2

=> VSO2 = VO2 = 2,8 (l)

a.b.c.

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25mol\)

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

0,25   0,125           0,25  ( mol )

\(V_{O_2}=n.22,4=0,125.22,4=2,8l\)

\(m_{H_2O}=n.M=0,25.18=4,5g\)

d.

\(S+O_2\rightarrow\left(t^o\right)SO_2\)

      0,125         0,125 ( mol )

\(V_{SO_2}=n.22,4=0,125.22,4=2,8l\)