Question

Answer 1

Bài 1:

Ta có: \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PT: \(2Ca+O_2\rightarrow2CaO\)

_________0,2____0,4 (mol)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

_0,4____________0,4 (mol)

⇒ m_Ca(OH)2 = 0,4.74 = 29,6 (g)

Bài 2:

a, Ta có: \(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

_____0,4___________0,6 (mol)

\(\Rightarrow m_{KClO_3}=0,4.122,5=49\left(g\right)\)

b, PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

_____1,2____0,6 (mol)

\(\Rightarrow V_{H_2}=1,2.22,4=26,88\left(l\right)\)

Bạn tham khảo nhé!

Answer 2

ui đọc lé con mắt á