Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 16.9 g of butane is mixed with 39. g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Round your answer to 2 significant digits.
To calculate the minimum mass of butane that could be left over in the reaction, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed when the reaction goes to completion and determines the amount of product that can be formed.
To find the limiting reactant, we compare the number of moles of each reactant to their respective stoichiometric coefficients in the balanced chemical equation. The balanced equation for the reaction is:
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
First, we need to convert the masses of butane and oxygen to moles:
Molar mass of butane (C4H10) = 58.12 g/mol
Molar mass of oxygen (O2) = 32.00 g/mol
Number of moles of butane = 16.9 g / 58.12 g/mol ≈ 0.29 mol
Number of moles of oxygen = 39. g / 32.00 g/mol ≈ 1.22 mol
Now, we need to compare the ratios of the moles of butane and oxygen to their stoichiometric coefficients:
Moles of butane : Moles of oxygen = 0.29 mol : 1.22 mol
From the balanced equation, we see that the stoichiometric coefficient of butane is 2 and the stoichiometric coefficient of oxygen is 13. Therefore, we can calculate the moles of butane needed to react completely with the given moles of oxygen:
Moles of butane needed = (13 mol of butane / 2 mol of oxygen) * 1.22 mol of oxygen
≈ 7.93 mol
Since the moles of butane available (0.29 mol) is less than the moles needed (7.93 mol), butane is the limiting reactant. This means that all of the butane will be consumed in the reaction, and there will be no butane left over.
Therefore, the minimum mass of butane that could be left over is 0 g (rounded to 2 significant digits).