Oblicz ile gramów azotu reaguje z tlenem w celu otrzymania 500g tlenku azotu (V).
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2N₂ + 5O₂ ---> 2N₂O₅
m2N₂ = 4 * 14u = 56u
m2N₂O₅ = 4 * 14u + 10 * 16u = 56u + 160u = 216u
56u ------------------216u
x ------------------- 500g
x = 56u * 500g / 216u
x = 129,6 g N₂
Z tlenem będzie reagowało129,6g azotu