\(m_{CO_2}=12+16\cdot2=44\)

Số mol của CO₂

\(n=\frac{m}{M}=\frac{11}{44}=0,25\)(mol)

Số mol của H₂

\(n=\frac{sophantu}{6.10^{23}}=\frac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

\(\Rightarrow\) V_H2(đktc) =n.22,4=1,5.22,4=33,6(lít)

=> V_H2(đktc) = 1,5 x 22,4 = 33,6 lít

 mCO2 = 12 + 16.2 = 44
Số mol của CO2
n = m/M = 11/44 = 0,25(mol)
Số mol của H2
n = số phần tử/6.10^23 = 9.10^23/6.10^23 = 1,5(mol)
=> VH2(đktc) = n.22,4 = 1,5.22,4 = 33,6(lít)

Ta có:

+ \(M_{CO_2}=12+16.2=44\) g/mol

⇒ \(n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=\dfrac{1}{4}mol\)

+ \(n=\dfrac{sophantu}{6.10^{23}}=\dfrac{9.10^{23}}{6.10^{23}}=\dfrac{3}{2}=mol\)

\(V_{H_2\left(đktc\right)}=n.22,4=\dfrac{3}{2}.22.4=33,6\left(l\right)\)

\(a,n_{CO_2}=\dfrac{m_{CO_2}}{M_{CO_2}}=\dfrac{11}{44}=0,25\left(mol\right)\\ b,n_{H_2}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\\ V_{H_2\left(đktc\right)}=1,5.22,4=33,6\left(l\right)\)

\(a.\)

\(n_{CO_2}=\dfrac{11}{44}=0.25\left(mol\right)\)

\(b.\)

\(n_{H_2}=\dfrac{9\cdot10^{23}}{6\cdot10^{23}}=1.5\left(mol\right)\)

\(V_{H_2}=1.5\cdot22.4=33.6\left(l\right)\)

Vấn đề em hỏi là gì nhỉ ?

Í bạn là 6,022 . 10^{23 ?}

\(a)n_{CO_2}=\dfrac{11}{44}=0,25\left(mol\right)\)

\(b)n_{H_2}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

\(\Rightarrow V_{H_2}=1,5.22,4=33,6\left(l\right)\)

a) $n_{CO_2} = \dfrac{11}{44} = 0,25(mol)$

b) $n_{H_2} = \dfrac{9.10^{23}}{6.10^{23}} = 1,5(mol)$

$V_{H_2} = 1,5.22,4 = 33,6(lít)$
c) $n_{SO_2} = \dfrac{5,6}{64} = 0,0875(mol)$

$V_{SO_2} = 0,0875.22,4 = 1,96(lít)$

Câu1) n_CO2 =m/M=11/44=0,25(mol)

         nH₂= 9.10²³/6.10²³=1,5(mol)

     V_H2 =n.22,4=1,5.22,4=33,6(l)

a) 

\(n_{SO_2}=\dfrac{m}{M}=\dfrac{3,2}{64}=0,05\left(mol\right)\\ n_{CO_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

b) 

\(n_{Cl_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\Rightarrow m_{Cl_2}=n.M=0,06.71=4,26\left(mol\right)\\ n_{Na_2CO_3}=n.M=0,5.106=53\left(g\right)\)

c) 

\(V_{N_2\left(\text{đ}ktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\\ n_{O_2}=\dfrac{m}{M}=\dfrac{4,8}{32}=0,15\left(mol\right)\Rightarrow V_{O_2\left(\text{đ}ktc\right)}=n.22,4=0,15.22,4=3,36\left(l\right)\)

V C O 2 = n C O 2  . 22,4 = 0,25.22,4 = 5,6 (l)

V O 2 = n O 2 .22,4 = 0,25.22,4 = 5,6 (l)

n N 2 == 0,75(mol)

→ V N 2 = n N 2 .22,4 = 0,75.22,4= 16,8 (l)

n C O 2 == 0,2 (mol)

→ V C O 2 = n C O 2 . 22,4 = 0,2.22,4 = 4,48 (l)

n H 2 ==1,5(mol)

→ V H 2 = n H 2 . 22,4 = 1,5. 22,4 = 33,6 (l)

n C O == 0,05(mol)

→ V C O = n C O . 22,4 = 0,05. 22,4 = 1,12 (l)