Question

Thể tích (đktc) của 2,2 g CO₂; của 4,8 g O₂

Answer 1

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Answer 2

\(n_{CO_2}=\dfrac{m}{M}=\dfrac{2,2}{44}=0,05\left(mol\right)\\ V_{O_2}=n.22,4=0,05.22,4=1,12\left(l\right)\\ n_{O_2}=\dfrac{m}{M}=\dfrac{4,8}{32}=0,15\left(mol\right)\\ V_{O_2}=n.22,4=0,15.22,4=3,36\left(l\right)\)

Answer 3

\(V_{CO_2}=\left(\dfrac{2,2}{44}\right).22,4=1,12l\)

\(V_{O_2}=\left(\dfrac{4,8}{32}\right).22,4=3,36l\)