To determine the amount of carbon dioxide (CO2) and water (H2O) formed when 5.0 moles of butane (C4H10) is completely combusted, we need to examine the balanced chemical equation for the combustion reaction of butane:
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
From the equation, we can see that for every 2 moles of butane, 8 moles of carbon dioxide and 10 moles of water are formed. Therefore, we can set up a proportion to calculate the moles of carbon dioxide and water formed:
(8 moles CO2 / 2 moles C4H10) = (10 moles H2O / 2 moles C4H10)
Using this proportion, we can calculate the moles of carbon dioxide and water formed:
Moles of CO2 = (8 moles CO2 / 2 moles C4H10) * 5.0 moles C4H10 = 20 moles CO2 Moles of H2O = (10 moles H2O / 2 moles C4H10) * 5.0 moles C4H10 = 25 moles H2O
To convert these moles to grams, we need to use the molar masses of carbon dioxide and water:
Molar mass of CO2 = 44.01 g/mol Molar mass of H2O = 18.015 g/mol
Grams of CO2 = Moles of CO2 * Molar mass of CO2 = 20 moles CO2 * 44.01 g/mol = 880.2 g CO2
Grams of H2O = Moles of H2O * Molar mass of H2O = 25 moles H2O * 18.015 g/mol = 450.4 g H2O
Therefore, when 5.0 moles of butane is completely combusted, 880.2 grams of carbon dioxide and 450.4 grams of water will be formed.