A

A

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,1           0,2                     0,1 
\(V_{H_2}=0,1.22,4=2,24l\\ m_{HCl}=\dfrac{0,2.36,5}{10}.100=73g\)

hỏi 1 lần thôi bn :v

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

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

Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b, Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

nH₂=6.72/22,4=0,3(mol)
PTHH: 2Al +   6HCl --> 2AlCl₃ + 3H₂
bài ra:  0,2 <-- 0,6  <--   0,2   <-- 0,3  /mol
a) mHCl = 0,6.36,5=21,9(g)
b) mAl = 0.2.24 = 4,8(g)
 

\(PTHH:2Al+6HCl->2AlCl_3+3H_2\)

              0,2<--0,6<----------0,2<------0,3    (mol)

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

\(m_{HCl}=n\cdot M=0,6\cdot\left(1+35,5\right)=21,9\left(g\right)\)

\(m_{AlCl_3}=n\cdot M=0,2\cdot\left(27+35,5\cdot3\right)=26,7\left(g\right)\)

a, PT: 

Ta có: nAl=23nH2=0,2(mol)

⇒mAl=0,2.27=5,4(g)

a, PT: 

Ta có: nAl=23nH2=0,2(mol)

⇒mAl=0,2.27=5,4(g)

Mg+Cl₂-to>MgCl₂

0,25---0,25----0,25 mol

n _Mg=6\24=0,25 mol

m _MgCl2=0,25.95=23,75g

V_Cl2=0,25.22,4=5,6l

a) PTHH : Mg + Cl₂ -> MgCl₂

\(n_{Mg}=\dfrac{m}{M}=\dfrac{6}{24}=0,25\left(mol\right)\)

b) \(m_{MgCl}=59,5.0,25=14,875\left(g\right)\)

c) \(V_{Cl}=0,25.22,4=5,6\left(l\right)\)

\(a,PTHH:Mg+Cl_2\underrightarrow{t^o}MgCl_2\)

\(b,n_{Mg}=\dfrac{m}{M}=\dfrac{6}{24}=0,25\left(mol\right)\\ Theo.PTHH:n_{Cl_2}=n_{MgCl_2}=n_{Mg}=0,25\left(mol\right)\\ m_{MgCl_2}=n.M=0,25.95=23,75\left(g\right)\\ c,V_{Cl_2\left(đktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\)

a) Mg + Cl₂ --t^o--> MgCl₂ 

Tỉ lệ Mg : Cl₂ = 1:1

b) Theo ĐLBTKL:

\(m_{Mg}+m_{Cl_2}=m_{MgCl_2}\)

=> \(m_{Cl_2}=19-4,8=14,2\left(g\right)\)

=> \(n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\)

=> \(V_{Cl_2}=0,2.22,4=4,48\left(l\right)\)

c) \(n_{MgCl_2}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

PTHH: Mg + Cl₂ --t^o--> MgCl₂ 

______0,5<--0,5<--------0,5__________(mol)

=> Số nguyên tử Mg = 0,5.6.10²³ = 3.10²³

=> Số phân tử Cl₂ = 0,5.6.10²³ = 3.10²³

Mg+2CH3COOH->(CH3COO)2Mg +H2

0,2------0,4--------------------0,2------------------0,2

nH2=0,2 mol

mMg=0,2.24=4,8g

NaOH+CH3COOH->CH3COONa+H2O

0,4-----------0,4 

Vdd=0,4/0,5=0,8l

Bài 1:

PTHH: \(CH_4+Cl_2\underrightarrow{a/s}CH_3Cl+HCl\)

Theo PTHH: \(n_{Cl_2}=n_{CH_3Cl}=n_{CH_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CH_3Cl}=0,15\cdot50,5=7,575\left(g\right)\\V_{CH_4}=V_{Cl_2}=3,36\left(l\right)\end{matrix}\right.\) 

Bài 2:

PTHH: \(CH_4+2O_2\underrightarrow{a/s}CO_2+2H_2O\)

Theo PTHH: \(\left\{{}\begin{matrix}n_{CO_2}=n_{CH_4}\\n_{O_2}=2n_{CH_4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=V_{CH_4}=5,6\left(l\right)\\V_{CO_2}=2V_{CH_4}=11,2\left(l\right)\end{matrix}\right.\)

C

8g

13C
14 8g ạ