a. Tính khối lượng: 0.3 mol Na2SO3; 247,9 lít khí CO2 đkc, 0.05 mol CaCl2 b. Tính số mol: 1.8 x 1023 phân tử NaCl, 7,437 l NO, 13g Zn c. Tính thể tích ở đkc: 8,8g CO2; 7,2g O2; 0.6g H2

$a,m_{Na_2SO_3}=n.M=0,3.126=37,8(g)\\ n_{CO_2}=\dfrac{247,9}{24,79}=10(mol)\Rightarrow m_{CO_2}=n.M=10.44=440(g)\\ m_{CaCl_2}=n.M=0,05.111=5,55(g)\\ b,n_{NaCl}=\dfrac{1,8.10^{23}}{6.10^{23}}=0,3(mol)\\ n_{NO}=\dfrac{7,437}{24,79}=0,3(mol)\\ n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2(mol)\\ c,n_{CO_2}=\dfrac{m}{M}=\dfrac{8,8}{44}=0,2(mol)\Rightarrow V_{CO_2}=0,2.24,79=4,958(l)\\ n_{O_2}=\dfrac{m}{M}=\dfrac{7,2}{32}=0,225(mol)\Rightarrow V_{O_2}=0,225.24,79=5,57775(l)\\ n_{H_2}=\dfrac{0,6}{2}=0,3(mol)\Rightarrow V_{H_2}=0,3.24,79=7,437(l)$

`a)` 

`m_{Na_2SO_3}= 126 . 0,3= 37,8(g)` 

`n_{CO_2}={247,9}/{24,70}= 10(mol)` 

`m_{CO_2}= 10. 44= 440(g)` 

`m_{CaCl_2}= 0,05. 111=5,55(g)` 

`b)` 

`n_{NaCl}={1,8. 10^{23}}/{6.10^{23}}= 0,3(mol)`

`n_{NO}={7,437}/{24,79}= 0,3(mol)`

`n_{Zn}={13}/{65}= 0,2(mol)`

`c)` 

`n_{CO_2}={8,8}/{44}= 0,2(mol)` 

`V_{CO_2}= 0,2. 24,79=4,958(l)` 

`n_{O_2}={7,2}/{32}=0,225(mol)` 

`V_{O_2}= 0,225. 24,79= 5,57775(l)` 

`n_{H_2}={0,6}/2= 0,3(mol)` 

`V_{H_2}= 0,3. 24,79=7,437(l)`