Question

a,3mol CO2(₫ktc)= ? l

b, 88g CO2(₫ktc)=? l

c, 3 mol H2O (₫ktc) ? g

Answer 1

a,3mol CO2(₫ktc)
=> \(V_{CO2}=22,4\times3=6,72\left(l\right)\)

b, 88g CO2(₫ktc)
=> \(n_{CO2}=\dfrac{88}{44}=2\left(mol\right)=>V_{CO2}=22,4\times2=44,8\left(l\right)\)

c, 3 mol H2O (₫ktc)
=> \(m_{H2O}=3\times18=54\left(g\right)\)

Answer 2

a) \(V_{CO_2}=3\times22,4=67,2\left(l\right)\)

b) \(n_{CO_2}=\dfrac{88}{44}=2\left(mol\right)\)

\(\Rightarrow V_{CO_2}=2\times22,4=44,8\left(l\right)\)

c) \(m_{H_2O}=3\times18=54\left(g\right)\)

Answer 3

a,67,2

b,44,8

c,54

Answer 4

a,67,2 l

b,44,8 l

c,54 g

Answer 5

a) Ta có : V_CO2 = n . 22,4 = 3 . 22,4 = 67,2 (l)

b) Ta có : n_CO2 = \(\dfrac{m}{M}\) = \(\dfrac{88}{44}\) = 2 (mol)

⇒ V_CO2 = n . 22,4 = 2 . 22,4 = 44,8 (l)

c) Ta có : m_H2O = n . M = 3 . 18 = 54 (g)

Chúc bạn học tốt !!!