2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
(44.0 CO2/1 mol CO2) (8 mol CO2/2 mol C4H10) (1 mol C4H10) (2.43 g C4H10) = 7.36 g CO2
Kirsten H.
asked 09/08/20Disposable lighters contain butane (C4H10) and produce CO2 and H2O.
Part 1-
Complete and balance the chemical equation for this combustion reaction. Do not add states of matter.
C4H10 + O2 —> ?
Part 2-
Determine how many grams of CO2 are produced by burning 2.43 g of C4H10.
___ g carbon dioxide.
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
(44.0 CO2/1 mol CO2) (8 mol CO2/2 mol C4H10) (1 mol C4H10) (2.43 g C4H10) = 7.36 g CO2
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