Step 1
The reaction must be completed and balanced as follows:
C3H8 + 5 O2 => 3 CO2 + 4 H2O
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Step 2
Information provided:
20.0 g of C3H8
10.0 g of O2
----
Information needed:
1 mole of C3H8 = 44.1 g
1 mole of O2 = 32.0 g
(use your periodic table please)
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Step 3
By stoichiometry,
C3H8 + 5 O2 => 3 CO2 + 4 H2O
44.1 g C3H8 ---------- 5 x 32.0 g O2
20.0 g C3H8 ---------- X
X = 20.0 g C3H8 x 5 x 32.0 g O2/44.1 g C3H8
X = 72.6 g O2
For 20.0 g of C3H8, 72.6 g of O2 is needed, but there is only 10.0 g of O2.
Therefore,
Answer:
The limiting reactant = O2
The excess = C3H8
