\(nHCl=0,2.0,3=0,06\\ 2Al+6HCl=>2AlCl3+3H2\\ =>nAl=0,02\left(mol\right)\\ =>mAl=0,02.27=0,54\left(g\right)\\ tacónAlCl3=0,02\left(mol\right)\\ =>Cm\left(AlCl3\right)=\dfrac{0,02}{0,2}=0,1\left(M\right)\)

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, \(n_{HCl}=0,2.0,3=0,06\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{1}{3}n_{HCl}=0,02\left(mol\right)\)

\(\Rightarrow m_{Al}=0,02.27=0,54\left(g\right)\)

c, \(n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,02\left(mol\right)\)

\(\Rightarrow C_{M_{AlCl_3}}=\dfrac{0,02}{0,2}=0,1\left(M\right)\)

nCO2=0,4(mol)

a) PTHH: 2 NaOH + CO2 -> Na2CO3 + H2O

0,8_________0,4________0,4(mol)

=> mNaOH=0,8.40=32(g) 

=>C%ddNaOH=(32/200).100=16%

b) mddNa2CO3=mddNaOH+mCO2=200+0,4.44=217,6(g)

mNa2CO3=106.0,4=42,4(g)

=>C%ddNa2CO3=(42,4/217,6).100=19,485%

Chúc em học tốt!

n_CO2=8,96/22,4=0,4mol

a/ CO₂+2NaOH→Na₂CO₃+H₂O

   0,4      0,8           0,4         0,4

m_NaOH=0,8.40=32g

C%dd_NaOH=m_ct/_mdd.100%=32/200.100%=16%

b/m_CO2=0,4.44=17,6g

Theo định luật bảo toàn khối lượng:

m_CO2+m_NaOH=m_Na2CO3

17,6g+200g=217,6g

m_Na2CO3=0,4.106=42,4g

C%ddNa2CO3=m_ct/mdd.100%=42,4/217,6.100=19,4852g

a) Fe + 2HCl --> FeCl₂ + H₂

b) n_HCl = 0,2.1 = 0,2 (mol)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

                       0,2---->0,1--->0,1

=> m_FeCl2 = 0,1.127 = 12,7 (g)

c) V_H2 = 0,1.22,4 = 2,24 (l)

a) Zn + 2HCl --> ZnCl₂ + H₂

b) n_HCl = 0,2.1 = 0,2 (mol)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

                      0,2----->0,1---->0,1

=> m_ZnCl2 = 0,1.136 = 13,6 (g)

c) V_H2 = 0,1.22,4 = 2,24 (l)

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

PTHH: 2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O

Mol:      0,4                         0,2        0,2

b) \(C\%_{ddHCl}=\dfrac{0,4.36,5.100\%}{100}=14,6\%\)

c) \(m_{CaCl_2}=0,2.101=20,2\left(g\right)\)

\(a,n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1mol\\ Na_2SO_3+2HCl\rightarrow2NaCl+SO_2+H_2O\\ n_{SO_2}=n_{Na_2SO_4}=0,1mol\\ V_{SO_2}=0,1.22,4=2,24l\\ b,n_{HCl}=0,1.2=0,2mol\\ C_{M_{HCl}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\\ c,n_{NaOH}=\dfrac{40.10}{100.40}=0,1mol\\ T=\dfrac{0,1}{0,1}=1\\ \Rightarrow Tạo,NaHSO_3\\ NaOH+SO_2\rightarrow NaHSO_3\\ m_{NaHSO_3}=0,1.64+0,1.40=10,4g\)

PTHH: \(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)

Ta có: \(n_{NaOH}=0,4\cdot2=0,8\left(mol\right)\)

\(\Rightarrow n_{SO_2}=0,4\left(mol\right)=n_{Na_2SO_3}\) \(\Rightarrow\left\{{}\begin{matrix}V_{SO_2}=0,4\cdot22,4=8,96\left(l\right)\\m_{Na_2SO_3}=0,4\cdot126=50,4\left(g\right)\\C_{M_{Na_2SO_3}}=\dfrac{0,4}{0,4}=1\left(M\right)\end{matrix}\right.\) 

Sau phản ứng thu được O2?

sao vậy ạ?

a) Zn + 2HCl → ZnCl2 + H2

b) Theo PTHH :

n H2 = n Zn = 13/65 = 0,2 (mol)

=> V H2 = 0,2.22,4 = 4,48(lít)

c) n ZnCl2 = n Zn = 0,2(mol)

=> m ZnCl2 = 0,2.136 = 27,2(gam)

d) CM ZnCl2 = 0,2/0,2 = 1M