Question #b5e69

Question

1540 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-14T16:15:53

$3.01 xx 10^21" molecules O"_2$

$M_(CO_2) = "220 mg"$
$MW_(CO_2) = "44.01 g/mol"$
$"O"_2 " molecules = ?"$

$"moles " CO_2 = (M_(CO_2))/(MW_(CO_2)) = ("220 mg"/"1000 mg/g")/"44.01 g/mol" = "0.00500 mol"$

$"moles O"_2 " per mole CO"_2 = 1$

$"Avogadro's number" = N_A = 6.022 xx 10^23" molecules/mol"$

$"O"_2 " molecules" = "moles CO"_2 xx "moles O"_2 /"moles CO"_2 xx N_A$

$= "0.00500 mol CO"_2 xx ("1 mol O"_2)/("1 mol CO"_2) xx (6.022 xx10^23 " molecules O"_2)/("1 mol O"_2)$

$= 3.01 xx 10^21" molecules O"_2$