A mixture of 50.00 grams propane, C3H8, and 100.00 grams oxygen, O₂, is combusted. C3H8 - 44.10 g/mol and the reaction is C3H8 (g) +502(g) → 3CO2(g) + 4H₂O(g). Select words from the dropdown menus to fill in the blanks in the statement below. Oxygen is the limiting reactant and is in excess. 22.42 grams of propane

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Chapter1: Chemical Foundations
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the answer is correct I just need the calculations please 

A mixture of 50.00 grams propane, C3Hs, and 100.00 grams oxygen, O₂, is combusted. C3H8 - 44.10
g/mol and the reaction is C3H8 (g) +50O2(g) → 3CO₂(g) + 4H₂O(g). Select words from the
dropdown menus to fill in the blanks in the statement below.
is the limiting reactant and
Oxygen
is in excess.
✓ 22.42
grams of
propane
Transcribed Image Text:A mixture of 50.00 grams propane, C3Hs, and 100.00 grams oxygen, O₂, is combusted. C3H8 - 44.10 g/mol and the reaction is C3H8 (g) +50O2(g) → 3CO₂(g) + 4H₂O(g). Select words from the dropdown menus to fill in the blanks in the statement below. is the limiting reactant and Oxygen is in excess. ✓ 22.42 grams of propane
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