2C2H2+5O2→2H2O+4CO2

4FeS2+11O2→2Fe2O3+8SO2

CaCO3+2HNO3→Ca(NO3)2+H2O+CO2

2HCl+Na2CO3→H2O+2NaCl+CO2

2H2S+O2→2H2O+2S

3AgNO3+H3PO4→3HNO3+Ag3PO4

a) C₂H₂  +  5O₂   --------> 2CO₂   +  H₂O

 4 FeS_{2   +}  11O₂       -------->8SO_{2  +}  2Fe₂O₃

CaCO3    +   2HNO₃ -------->Ca(NO₃)₂  +  CO₂  +H₂O

Na₂CO₃    +2HCl  ------>2NaCl   +CO₂    +  H₂O

2H₂S    +  3O₂  ------>  2SO₂  +  2 H₂O

3AgNO₃  +  H₃PO_{4  ------->}Ag₃PO_{4  +}  3HNO₃

a,2SO2+O2--->2SO3

b,Al+HCl---->KOH+H2 ( cái này bạn vt sai r)

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

c,3Ca+2H3 PO4--->3H2+Ca3 (PO4)2

d,2C6H6+9O2---> 6CO2+6H2O

e,C2H6O+3O2--->2CO2+3H2O

f,  CxHyOz + (x+y/4-z/2) O2 ---> xCO2+ y/2 H2O 

giải thích ra hộ mình nha

\(n_{HCl}=0.3\left(mol\right)\)

\(n_{K_2CO_3}=0.12\cdot0.6=0.072\left(mol\right)\)

\(n_{NaHCO_3}=0.12\left(mol\right)\)

\(K_2CO_3+HCl\rightarrow KHCO_3+KCl\)

\(0.072......0.072..........0.072\)

\(KHCO_3+HCl\rightarrow KCl+CO_2+H_2O\)

\(0.072.......0.072.....................0.072\)

\(n_{HCl\left(cl\right)}=0.3-0.072-0.072=0.156\left(mol\right)\)

\(NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O\)

\(0.12............0.12..................0.12\)

\(\Rightarrow HCldư\)

\(n_{CO_2}=0.12+0.072=0.192\left(mol\right)\)

PTHH: \(CaO+2HCl\rightarrow CaCl_2+H_2O\)  (1)

            \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)  (2)

a) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{CaCl_2}=\dfrac{33,3}{111}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{CaO}=0,1mol\\n_{CaCO_3}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaO}=0,1\cdot56=5,6\left(g\right)\\m_{CaCO_3}=0,2\cdot100=20\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{hh}=20+5,6=25,6\left(g\right)\)

b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=0,2mol\\n_{HCl\left(2\right)}=0,4mol\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{HCl}=0,6mol\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,3}=2\left(M\right)\)

PT: \(CaO+2HCl\rightarrow CaCl_2+H_2O\) (1)

\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\) (2)

a, Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(\Sigma n_{CaCl_2}=\dfrac{33,3}{111}=0,3\left(mol\right)\)

Theo PT (2): \(n_{CaCl_2}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow n_{CaCl_2\left(1\right)}=0,3-0,2=0,1\left(mol\right)\)

Theo PT (1): \(n_{CaO}=n_{CaCl_2}=0,1\left(mol\right)\)

Theo PT (2): \(n_{CaCO_3}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_A=m_{CaO}+m_{CaCO_3}=0,1.56+0,2.100=25,6\left(g\right)\)

b, Theo PT (1) + (2): \(\Sigma n_{HCl}=2n_{CaO}+2n_{CaCO_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,6}{0,3}=2M\)

Bạn tham khảo nhé!

\(SO_2\): Lưu huỳnh đioxit, oxit axit

\(HCl\): Axit clohiđric, axit

\(CaO\): Canxi oxit, oxit bazơ

\(NaOH\): Natri hiđroxit, bazơ

\(H_2SO_4\): Axit sunfuric, axit

\(NaCl\): Natri clorua, muối trung hoà

\(Al\left(OH\right)_3\): Nhôm hiđroxit, lưỡng tính

\(SiO_2\): Silic đioxit, oxit axit

\(KNO_3\): Kali nitrat, muối trung hoà

\(CO\): Cacbon monoxit, oxit trung tính

\(H_3PO_4\): Axit photphoric, axit

\(NaHCO_3\): Natri hiđrocacbonat, muối axit

\(HNO_3\): Axit nitric, axit

\(CO_2\): Cacbon đioxit, oxit axit

\(Ca\left(H_2PO_4\right)_2\): Canxi đihiđrophotphat, muối axit

\(Ca\left(OH\right)_2\): Canxi hiđroxit, bazơ

a) \(CaO+SO_2\rightarrow CaSO_3\)

b) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

c) \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

d) \(Fe\left(OH\right)_2+H_2SO_4\rightarrow FeSO_4+2H_2O\)

e) \(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)

f) \(2HNO_3+CuO\rightarrow Cu\left(NO_3\right)_2+H_2O\)

 Chúc bạn học tốt

nHCl (ban đầu) = 0.6(mol)
gọi nK2CO3 = x (mol)
nNa2CO3 = y(mol)
Ta có : x + y = nCO2 = 0.25(mol)
K2CO3 + 2HCl ---> 2KCl + H2O + CO2
x________2x____________________x
Na2CO3 + 2HCl ---> 2NaCl + H2O + CO2
y_________2y____________________y
=> n(HCl pư) = 2 x 0.25 = 0.5(mol)
=> nHCl dư = 0.1(mol) = nNaOH
=> dd thu được gồm : K(+) , Na(+), Cl(-) , Na(+)
Ta có : 78x + 46y = 39.9 - 0.6*35.5 - 23*0.1 = 16.3
=> x = 0.15
y = 0.1

BẠN THAM KHẢO