Final answer:
To find the required amount of oxygen for the combustion of 47.2 grams of butane, stoichiometry from the balanced equation is used, resulting in 168.9 grams of oxygen needed.
Step-by-step explanation:
To determine how many grams of oxygen are required for the complete combustion of 47.2 g of butane (C4H10), we can use stoichiometry based on the balanced chemical equation:
2C4H10 (g) + 13O2 (g) → 8CO2 (g) + 10H2O (g)
First, we convert the mass of butane to moles using its molar mass (58.12 g/mol). Next, using the stoichiometry of the balanced equation, we find the moles of oxygen needed. Then, we convert the moles of oxygen to grams using the molar mass of oxygen (32.00 g/mol).
Calculating the moles of butane:
(47.2 g C4H10) × (1 mol C4H10 / 58.12 g C4H10) = 0.812 moles of C4H10
Next, using the stoichiometric relationship:
(0.812 moles C4H10) × (13 moles O2 / 2 moles C4H10) = 5.278 moles of O2
Then, calculate the mass of oxygen:
(5.278 moles O2) × (32.00 g O2 / 1 mol O2) = 168.896 g of O2
Therefore, 168.9 grams of oxygen are required for the complete combustion of 47.2 g of butane.