Question

đốt cháy hoàn toàn 13 gam magie thì cần bao nhiêu lít oxi (đktc)

Answer 1

\(n_{Mg}=\dfrac{13}{24}mol\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

\(\dfrac{13}{24}\)      \(\dfrac{13}{48}\)                          ( mol )

\(V_{O_2}=\dfrac{13}{48}.22,4=\dfrac{91}{15}l\)

Answer 2

2Mg+O2-to>2MgO

\(\dfrac{13}{24}\)---\(\dfrac{13}{48}\)

n Mg=\(\dfrac{13}{24}\)mol

=>VO2=\(\dfrac{13}{48}\).22,4=6,06l

Answer 3

nMg = 13 : 24 = 13/24 (mol) 
pthh : 2Mg + O2 -t--> 2MgO
          13/24->13/48 (mol) 
=> VO2 = 13/48 . 22,4 = 6,06 (l)