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19,5 g butane are needed.

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Francesca Feil

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2y ago
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10y ago

1.02g C4H10 /58.12 g/mol C4H10= .0175 mol C4H10

.0175 mol C4H10 * 10 mol H2O / 2 mol C4H10 = .0877 mol H2O

.0877 mol H2O * 18.02 g H2O = 1.58 g H20

Final Answer: 1.58 g H2O

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14y ago

Determine the mass of carbon dioxide produced when 0.85 g of butane reacts with oxygen according to the following equation 2 C4H10 13 O2 8 CO2 10 H2O?

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8y ago

The reaction is:
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
The answer is 24 moles.

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Q: How many moles of carbon dioxide (CO2) are produced when reacting 6.00 moles of butane (C4H10) in excess oxygen (O2)?
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