Question

B1: Tính số mol của

a) 2,8 gam Fe

b) 10,2 gam Al₂O₃

c) 2,24 lít H₂

d) 6,72 lít CO₂

Answer 1

n_Fe=\(\frac{2,8}{56}=0,05\)(mol)

\(n_{Al_2O_3}=\frac{10,2}{102}=0,1\)(mol)

\(n_{H_2}=\frac{2,24}{22,4}=0,1\)(mol)

\(n_{CO_2}=\frac{6,72}{22,4}=0,3\)(mol)

Answer 2

a) \(n_{Fe}=\frac{2,8}{56}=0,05\left(mol\right)\)

b) \(n_{Al_2O_3}=\frac{10,2}{102}=0,1\left(mol\right)\)

c) \(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)

d) \(n_{CO_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)

Answer 3

a) n_Fe=\(\frac{m}{M}\)=\(\frac{2,8}{56}\)=0,05 (mol)

b)n_Al2O3=\(\frac{m}{M}\)=\(\frac{10,2}{102}\)=0,1 (mol)

c)n_H2=\(\frac{V}{22,4}\)=\(\frac{2,24}{22,4}\)=0,1 (mol)

d)n_CO2=\(\frac{V}{22,4}\)=\(\frac{6,72}{22,4}\)=0,3 (mol)