Calculate mass of carbon dioxide (CO2) formed when 87 g of butane (C4H10) is burned completely in excess of oxygen.Give your answer with the appropriate unit.

To solve this problem, we need to use the concept of stoichiometry in chemistry. Here are the steps:

1. Write down the balanced chemical equation for the combustion of butane:

   2C4H10 + 13O2 -> 8CO2 + 10H2O

   This equation tells us that 2 moles of butane react with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.

2. Calculate the molar mass of butane (C4H10). The atomic masses of carbon (C) and hydrogen (H) are approximately 12 g/mol and 1 g/mol respectively. So, the molar mass of butane is:

   (412 g/mol) + (101 g/mol) = 58 g/mol

3. Calculate the number of moles of butane in 87 g.

   Moles = mass / molar mass = 87 g / 58 g/mol = 1.5 moles

4. From the balanced chemical equation, we know that 2 moles of butane produce 8 moles of CO2. So, 1.5 moles of butane will produce:

   (1.5 moles / 2) * 8 = 6 moles of CO2

5. Finally, calculate the mass of CO2 produced. The molar mass of CO2 (carbon dioxide) is (12 g/mol for C) + (2*16 g/mol for O) = 44 g/mol. So, the mass of CO2 produced is:

   Mass = moles * molar mass = 6 moles * 44 g/mol = 264 g

So, 87 g of butane will produce 264 g of carbon dioxide when burned completely in excess oxygen.