Question

Tính số mol của :

a/ 2,45g KClO3

b/ 2,04g Al2O3

Mn giúp mình nhanh với ạ mình đang cần gấp. Cảm ơn mn

Answer 1

a) M_KClO3=39+35,5+16.2=106,5 g/mol

\(n_{KClO_3}=\frac{m}{M}=\frac{2,45}{106,5}=0,02\) mol

b) M_Al2O3=27.2+16.3=102 g/mol

\(n_{Al_2O_3}=\frac{m}{M}=\frac{2,04}{102}=0,02\) mol

Answer 2

a, M KClO3 = 122,5g/mol

=> nKClO3= 2,45/ 122,5=0,02mol

b, M Al2O3= 102g/mol

=> nAl2O3= 2,04/ 101=0,02mol

Answer 3

a) n\(_{KClO_3}=\frac{2,45}{122.5}=0,02\left(mol\right)\)

b) n\(_{A_{ }l2O3}\frac{2,04}{102}=0,02\left(mol\right)\)

Chúc bạn học tốt

Answer 4

\(n_{KClO_3}=\frac{2,45}{122,5}=0,02\left(mol\right)\)

\(m_{Al_2O_3}=\frac{2,04}{102}=0,02\left(mol\right)\)

Answer 5

\(n_{KClO_3}=\frac{m}{M}=\frac{2.45}{122.5}=0.02\left(mol\right)\\ n_{Al_2O_3}=\frac{m}{M}=\frac{2.04}{102}=0.02\left(mol\right)\)

Answer 6

\(n=\frac{m}{M}\)

a) \(n_{KClO_3}=\frac{2,45}{122,5}=0,02\left(mol\right)\)

b) \(n_{Al_2O_3}=\frac{2,04}{102}0,02\left(mol\right)\)