Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 29.1 g of butane is mixed with 64. g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

limiting reactant is butane. .19299moles
1.69238 moles of oxygen gas left over

it appears to me you are definitely off on significant digits.

C4H10+6.5 O2 >>> 5H2O + 4CO2

you start with 29.1/58=.5 moles butane
and 2 moles of O2

for the butane of .5 moles, you need 6.5x.5=3.25 moles O2. You do not have that much, so the limiting reactant is O2.

So if all the O2 is expended, you only use 2/6.5 moles butane

amount of butane left=orig-used
= .5-2/6.5=.29 moles butane left over.

So check all that, it is certainly much different than what you got. I did most of the math in my head.

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

Rmm of 1 mole oxygen=32g

13 mole=416g

mass of 1 mole of butane=58g
2 mole=116g

116g C4H10 requires 416g O2

17.8 C4H10 will require 64g O2

excess butane=29.1-17.8=11.3g

To determine the minimum mass of butane that could be left over, we first need to calculate the moles of butane and oxygen used in the reaction.

1. Calculate the moles of butane:
Given mass of butane = 29.1 g
Molar mass of butane (C₄H₁₀) = 58.12 g/mol (from the periodic table)

Moles of butane = mass of butane / molar mass of butane
Moles of butane = 29.1 g / 58.12 g/mol
Moles of butane = 0.5 moles (rounded to one decimal place)

2. Calculate the moles of oxygen:
Given mass of oxygen = 64.0 g
Molar mass of oxygen (O₂) = 32.00 g/mol (from the periodic table)

Moles of oxygen = mass of oxygen / molar mass of oxygen
Moles of oxygen = 64.0 g / 32.00 g/mol
Moles of oxygen = 2.0 moles (rounded to one decimal place)

3. Determine the limiting reactant:
To find the limiting reactant, we compare the mole ratios of butane and oxygen in the balanced chemical equation. The balanced equation is as follows:

C₄H₁₀ (butane) + 13/2 O₂ (oxygen) → 4 CO₂ (carbon dioxide) + 5 H₂O (water)

The ratio of butane to oxygen is 1:13/2 or 2:13. This means that for every 2 moles of butane, we need 13/2 moles of oxygen.

Mole ratio of butane to oxygen = 2/13

Comparing this ratio with the actual moles of butane and oxygen calculated earlier:
Actual mole ratio of butane to oxygen = 0.5 moles / 2.0 moles = 0.25

Since the actual mole ratio is less than the stoichiometric ratio, butane is the limiting reactant.

4. Calculate the moles of carbon dioxide produced:
Using the mole ratio from the balanced equation, we can determine the moles of carbon dioxide produced.

Moles of carbon dioxide = Moles of butane × (4 moles of CO₂ / 2 moles of butane)
Moles of carbon dioxide = 0.5 moles × (4 / 2)
Moles of carbon dioxide = 1.0 moles

5. Calculate the moles of butane left over:
Since butane is the limiting reactant, the amount of moles of butane left over is determined by the stoichiometry of the reaction.

Moles of butane left over = Moles of butane initially - Moles of butane reacted
Moles of butane left over = 0.5 moles - 0.5 moles
Moles of butane left over = 0 moles

6. Calculate the mass of butane left over:
To calculate the mass of butane left over, we multiply the moles of butane left over by the molar mass of butane.

Mass of butane left over = Moles of butane left over × Molar mass of butane
Mass of butane left over = 0 moles × 58.12 g/mol
Mass of butane left over = 0 g

Therefore, the minimum mass of butane that could be left over is 0 g.

To find the minimum mass of butane that could be left over, we need to determine the limiting reactant in the chemical reaction.

First, we need to convert the given masses into moles using their respective molar masses:

- Butane (C4H10) has a molar mass of 58.12 g/mol.
- Oxygen (O2) has a molar mass of 32.00 g/mol.

Converting the masses to moles:

- Butane: 29.1 g / 58.12 g/mol = 0.50006 moles (rounded to the correct number of significant digits).
- Oxygen: 64.0 g / 32.00 g/mol = 2.000 moles.

Next, we need to determine the molar ratio of butane to oxygen in the balanced chemical equation. From the given reaction, we can see that the ratio is 1:2. This means that for every 1 mole of butane, we need 2 moles of oxygen.

Comparing the moles of each reactant:

- Butane: 0.50006 moles
- Oxygen: 2.000 moles

Since the molar ratio is 1:2, it tells us that 1 mole of butane requires 2 moles of oxygen. Therefore, the butane is the limiting reactant because we have more oxygen than needed.

Next, we need to calculate the moles of oxygen that reacted with the moles of butane:

- Moles of oxygen used in the reaction = 0.50006 moles of butane x 2 moles of oxygen/mole of butane = 1.00012 moles (rounded to the correct number of significant digits).

Now we can calculate the moles of oxygen left over:

- Moles of oxygen left over = 2.000 moles - 1.00012 moles = 0.99988 moles (rounded to the correct number of significant digits).

Finally, we can calculate the mass of butane left over using the moles of oxygen left over:

- Mass of butane left over = Moles of oxygen left over x molar mass of butane
- Mass of butane left over = 0.99988 moles x 58.12 g/mol = 58.11 g (rounded to the correct number of significant digits).

Therefore, the minimum mass of butane that could be left over after the chemical reaction is 58.11 g.