Phương trình hóa học CaCO3 → CaO + CO2.

a) nCaO =  = 0,2 mol.

Theo PTHH thì nCaCO3 = nCaO = 0,2 (mol)

b) nCaO =  = 0,125 (mol)

Theo PTHH thì nCaCO3 = nCaO = 0,125 (mol)

mCaCO3 = M.n = 100.0,125 = 12,5 (g)

c) Theo PTHH thì nCO2 = nCaCO3 = 3,5 (mol)

VCO2 = 22,4.n = 22,4.3,5 = 78,4 (lít)

d) nCO2 =  = 0,6 (mol)

Theo PTHH nCaO = nCaCO3 = nCO2 = 0,6 (mol)

mCaCO3 = n.M = 0,6.100 = 60 (g)

mCaO = n.M = 0,6.56 = 33,6 (g)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo pt: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2mol\)

Phương trình hóa học CaCO\(_3\) → CaO + CO\(_2\).

a) nCaO = \(\frac{11,2}{56}\) = 0,2 mol.

Theo PTHH thì nCaCO\(_3\) = nCaO = 0,2 (mol)

b) nCaO = \(\frac{7}{56}\) = 0,125 (mol)

Theo PTHH thì nCaCO\(_3\) = nCaO = 0,125 (mol)

mCaCO\(_3\) = M.n = 100.0,125 = 12,5 (g)

c) Theo PTHH thì nCO\(_2\) = nCaCO\(_3\) = 3,5 (mol)

VCO\(_2\) = 22,4.n = 22,4.3,5 = 78,4 (lít)

d) nCO\(_2\) = \(\frac{13,44}{22,4}\) = 0,6 (mol)

Theo PTHH nCaO = nCaCO\(_3\) = nCO\(_2\) = 0,6 (mol)

mCaCO\(_3\) = n.M = 0,6.100 = 60 (g)

mCaO = n.M = 0,6.56 = 33,6 (g)

                                   #shin

PTHH : CaCO₃ --> Cao + CO₂

a, n_CaO = \(\frac{11,2}{56}=0,2\left(mol\right)\)

Theo PTHH : n_CaO = n_CaCO3 = 0,2 mol

Vậy : Cần dùng 0,2 mol  CaCO₃ để điều chế đc 11,2 g CaO

b, Mấy câu sau dễ tự làm

a) 2Al + 6HCl --> 2AlCl₃ + 3H₂

b) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\); n_HCl = 0,5.2 = 1 (mol)

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{1}{6}\) => Al hết, HCl dư

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             0,1->0,3---->0,1---->0,15

=> V_H2 = 0,15.22,4 = 3,36 (l)

c) \(\left\{{}\begin{matrix}C_{M\left(AlCl_3\right)}=\dfrac{0,1}{0,5}=0,2M\\C_{M\left(HCl.dư\right)}=\dfrac{1-0,3}{0,5}=1,4M\end{matrix}\right.\)

a) \(n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Ta có: \(n_{CaCO_3}=n_{CaO}=0,2\left(mol\right)\)

b) \(n_{CaO}=\dfrac{7}{56}=0,125\left(mol\right)\)

Ta có :  \(n_{CaCO_3}=n_{CaO}=0,125\left(mol\right)\)

=> \(m_{CaCO_3}=0,125.100=12,5\left(g\right)\)

c) \(n_{CO_2}=n_{CaCO_3}=3,5\left(mol\right)\)

=> \(V_{CO_2}=3,5.22,4=78,4\left(lít\right)\)

d) \(n_{CO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Ta có: \(n_{CO_2}=n_{CaCO_3}=n_{CaO}=0,6\left(mol\right)\)

=> \(m_{CaCO_3}=0,6.100=60\left(g\right)\)

\(m_{CaO}=0,6.56=33,6\left(g\right)\)

2Al+6HCl->2AlCl3+3H2

0,5----1,5-------------------------0,75

n H2=0,075 mol

cần 10 quả bóng =>0,075.10=0,75 mol

=>mAl=0,5.27=13,5g

m HCl=1,5.36,5=54,75g

=>m dd=75g

\(n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\) 
           0,1125   0,15                     0,075 
\(m_{Al}=0,1125.27=3,0375\left(g\right)\\ m_{HCl}=\dfrac{\left(0,15.36,5\right).100}{7,3}=75\left(g\right)\) 

ta có n_CO2=\(\frac{13.44}{22.4}\)=0,6 mol

bt1) Fe₂O₃+ CO\(\rightarrow\) CO₂+Fe

ta có n_Fe= 0,6 mol

vậy m_Fe=0,6.56=33,6

bt2) ta có n_Fe=1,12:56=0,02 mol

PTHH: Fe203+3CO\(\rightarrow\)2Fe+CO2

                                  0,02\(\rightarrow\)0,01(mol)

V_CO= 0,01. 22,4=0,224(lít)

bạn ơi mink nhầm bài 2 mink làm phải là bài 3 mới đúng bạn nhé

a) PTHH: NaOH + Al + H2O -> NaAlO2 + 3/2 H2

b) nH2= 0,6(mol)

-> nAl=0,4(mol) => mAl=0,4.27=10,8(g)

c) nAl=0,18((mol); nNaOH=0,2(mol)

PTHH: 0,18/1 < 0,2/1

=> Al hết, NaOH dư, tính theo nAl.

-> nH2= 3/2. 0,18=0,27(mol)

=>V(H2,đktc)=0,27.22,4= 6,048(l)

\(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)

\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)

\(...........0.4.........................0.6\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

\(n_{Al}=\dfrac{4.86}{27}=0.18\left(mol\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0.2\left(mol\right)\)

\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)

\(2.................2\)

\(0.2...............0.18\)

\(LTL:\dfrac{0.2}{2}>\dfrac{0.18}{2}\)

\(\Rightarrow NaOHdư\)

\(n_{H_2}=0.18\cdot\dfrac{3}{2}=0.27\left(mol\right)\)

\(V_{H_2}=0.27\cdot22.4=6.048\left(l\right)\)

a, PTHH: 2NaOH + 2Al + 2H₂O → 2NaAlO₂ + 3H₂↑

b, nH2=0,6 mol -mAl=10,8 g 

c, nAl=0,16(592) mol 

   nNaOH=0,2 mol 

-NaOH hết, Al dư 

-VH2=6,72 l

\(\left(a\right)2Al+3H_2O\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \left(b\right)n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{Al}=\dfrac{0,6.2}{3}=0,4mol\\ m_{Al}=0,4.27=10,8g\\ \left(c\right)n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4Al+3O_2\underrightarrow{t^0}2Al_2O_3\\ \Rightarrow\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Al.dư\\ n_{Al_2O_3}=\dfrac{0,15.2}{3}=0,1mol\\ m_{oxit}=m_{Al_2O_3}=0,1.102=10,2g\)

a: \(2Al+3H_2SO_4\rightarrow1Al_2\left(SO_4\right)_3+3H_2\uparrow\)

    0,4           0,6                0,2             0,6

b: \(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)

=>\(n_{Al}=0.4\left(mol\right)\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

c: \(4Al+3O_2\rightarrow2Al_2O_3\)

0,4                      0,2

\(m_{Al_2O_3}=0.2\left(27\cdot2+16\cdot3\right)=0.2\cdot102=20.4\left(g\right)\)

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH ;

2Al + 3H2SO4 --> Al2(SO4)3 + 3H2↑

  0,4           0,6               0,2           0,6

4Al + 3O2 ---> 2Al2O3

0,2       0,15          0,1 

\(n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)

\(\dfrac{0,4}{4}>\dfrac{0,15}{3}\)

--> Tính theo oxi

\(b,m_{Al}=0,4.27=10,8\left(g\right)\)

\(c,m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)