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.

This is a limiting reagent problem with a twist; i.e., how much is left over and not how much something is formed. Here is a detailed and worked example of a limiting reagent problem. Just follow the steps.

http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html

To calculate the minimum mass of butane left over in the reaction, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

To find the limiting reactant, we can use the concept of stoichiometry and compare the amount of product that could be formed from each reactant.

The balanced chemical equation for the reaction is:
C4H10 + O2 -> CO2 + H2O

From the equation, we can see that the molar ratio of butane to carbon dioxide is 1:1. This means that 1 mole of butane reacts to produce 1 mole of carbon dioxide.

First, we need to convert the given masses of butane and oxygen to moles. The molar mass of butane (C4H10) is 58.12 g/mol, and the molar mass of oxygen (O2) is 32.00 g/mol.

Moles of butane = mass of butane / molar mass of butane
Moles of butane = 16.9 g / 58.12 g/mol
Moles of butane = 0.29 mol (rounded to 2 decimal places)

Moles of oxygen = mass of oxygen / molar mass of oxygen
Moles of oxygen = 39. g / 32.00 g/mol
Moles of oxygen = 1.22 mol (rounded to 2 decimal places)

Now, using the mole ratio from the balanced equation, we can determine the amount of carbon dioxide that could be formed from each reactant.

Amount of carbon dioxide from butane = moles of butane × 1
Amount of carbon dioxide from butane = 0.29 mol

Amount of carbon dioxide from oxygen = moles of oxygen × 1
Amount of carbon dioxide from oxygen = 1.22 mol

Since the stoichiometry ratio indicates that 1 mole of butane reacts to produce 1 mole of carbon dioxide, and the moles of oxygen (1.22 mol) are greater than the moles of butane (0.29 mol), we can conclude that oxygen is the limiting reactant.

To find the mass of butane that is left over, we need to calculate the amount of butane that reacts with the given amount of oxygen.

Amount of butane reacted = moles of oxygen × 1 (stoichiometry ratio)
Amount of butane reacted = 1.22 mol

Moles of butane left over = moles of butane - amount of butane reacted
Moles of butane left over = 0.29 mol - 1.22 mol
Moles of butane left over = -0.93 mol

Since we cannot have negative moles, this means that all the butane is consumed and there is no butane left over. Therefore, the minimum mass of butane left over is 0 g.

To determine the minimum mass of butane that could be left over, we need to compare the ratio of the amount of butane to the amount of oxygen used in the reaction.

1. Start by calculating the number of moles for each substance using their respective masses and molar masses.

The molar mass of butane (C4H10) is:
(4*12.01 g/mol) + (10*1.01 g/mol) = 58.12 g/mol

The number of moles of butane is:
16.9 g / 58.12 g/mol ≈ 0.291 moles of butane

The molar mass of oxygen (O2) is:
2*16.00 g/mol = 32.00 g/mol

The number of moles of oxygen is:
39. g / 32.00 g/mol ≈ 1.219 moles of oxygen

2. Write the balanced chemical equation for the reaction:
C4H10 + 13O2 -> 8CO2 + 10H2O

According to the balanced equation, the ratio of butane to oxygen is 1:13. This means that for every 1 mole of butane, we need 13 moles of oxygen.

3. Calculate the amount of butane that is needed to react with 1.219 moles of oxygen:
0.291 moles of butane / 1.219 moles of oxygen ≈ 0.239 moles of butane

4. Convert the moles of butane back to mass:
0.239 moles of butane * 58.12 g/mol = 13.9 g of butane

Therefore, the minimum mass of butane that could be left over is approximately 13.9 grams.