Determine the mass of carbon dioxide produced when 0.85 grams of butane (C4 H10 ) reacts with oxygen according to the following balanced chemical equation: 2 C4 H10 + 13 O2 → 8 CO2 + 10 H2 O

Step 1: Calculate the molar mass of butane (C4H10) The molar mass of butane is calculated by adding the molar masses of its constituent atoms. Carbon (C) has a molar mass of 12.01 g/mol, and hydrogen (H) has a molar mass of 1.008 g/mol. Therefore, the molar mass of butane is (412.01 g/mol) + (101.008 g/mol) = 58.12 g/mol.

Step 2: Calculate the number of moles of butane The number of moles of butane is calculated by dividing the mass of butane by its molar mass. Therefore, the number of moles of butane is 0.85 g / 58.12 g/mol = 0.0146 mol.

Step 3: Use the stoichiometry of the reaction to calculate the number of moles of carbon dioxide From the balanced chemical equation, we can see that 2 moles of butane produce 8 moles of carbon dioxide. Therefore, 0.0146 mol of butane will produce (0.0146 mol * 8) / 2 = 0.0584 mol of carbon dioxide.

Step 4: Calculate the mass of carbon dioxide The molar mass of carbon dioxide (CO2) is (112.01 g/mol) + (216.00 g/mol) = 44.01 g/mol. Therefore, the mass of carbon dioxide produced is 0.0584 mol * 44.01 g/mol = 2.57 g.

So, the mass of carbon dioxide produced when 0.85 grams of butane reacts with oxygen is approximately 2.57 grams.