Bài 1: Hãy tính số mol của: a. 13g Zn; 5,4g Al; 4,8g Mg; 4,9g H2SO4; 20g NaOH; 51g Al2O3; 45g Fe(OH)2; 171g Al2(SO4)3 b. Ở đktc của: 2,24 lít O2; 3,36 lít CO2; 67,2 lít SO2; 5,6 lít NO2; 13,44 lít NH3; 8,96 lít CH4;

$a)n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2(mol)$

$n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2(mol)$

$n_{Mg}=\dfrac{m}{M}=\dfrac{4,8}{24}=0,2(mol)$

$n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{4,9}{98}=0,05(mol)$

$n_{NaOH}=\dfrac{m}{M}=\dfrac{20}{40}=0,5(mol)$

$n_{Al_2O_3}=\dfrac{m}{M}=\dfrac{51}{102}=0,5(mol)$

$n_{Fe(OH)_2}=\dfrac{m}{M}=\dfrac{45}{90}=0,5(mol)$

$n_{Al_2(SO_4)_3}=\dfrac{m}{M}=\dfrac{171}{342}=0,5(mol)$

$b)n_{O_2}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1(mol)$

$n_{CO_2}=\dfrac{V}{22,4}=\dfrac{3,36}{22,4}=0,15(mol)$

$n_{SO_2}=\dfrac{V}{22,4}=\dfrac{67,2}{22,4}=3(mol)$

$n_{NO_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25(mol)$

$n_{NH_3}=\dfrac{V}{22,4}=\dfrac{13,44}{22,4}=0,6(mol)$

$n_{CH_4}=\dfrac{V}{22,4}=\dfrac{8,96}{22,4}=0,4(mol)$