Refer to the following unbalanced equation: C6H14 + O2 CO2 + H2O

What mass of oxygen (O2) is required to react completely with 10.8 g of C6H14

Well, let's break it down! We have the unbalanced equation C6H14 + O2 CO2 + H2O. According to this equation, we need 6 moles of O2 to react with 1 mole of C6H14.

To figure out the mass of O2 required to react with 10.8g of C6H14, we need to find the molar mass of C6H14, which is approximately 86g/mol. Now, we can set up a proportion to calculate the mass of O2 needed:

10.8g C6H14 / 86g C6H14 = xg O2 / (6 mol O2 * 32g/mol O2)

By solving this proportion, we can find the mass of O2 required. Let's do some math:

10.8g C6H14 * (6 mol O2 * 32g/mol O2) / 86g C6H14 = xg O2

After the calculations, we find that approximately 15.6g of O2 is required to react completely with 10.8g of C6H14. So, you might want to have some extra oxygen on hand, just in case!

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Let's balance it:

C6H14 + (19/2) O2 → 6 CO2 + 7 H2O

Now that the equation is balanced, we can use it to calculate the mass of oxygen required to react completely with 10.8 g of C6H14.

Step 1: Calculate the molar mass of C6H14
C6H14 has a molar mass of:
(6 × atomic mass of carbon) + (14 × atomic mass of hydrogen)
= (6 × 12.01 g/mol) + (14 × 1.01 g/mol)
= 72.06 g/mol + 14.14 g/mol
= 86.20 g/mol (approximately)

Step 2: Calculate the number of moles of C6H14
Number of moles = Mass / Molar mass
Number of moles of C6H14 = 10.8 g / 86.20 g/mol
≈ 0.125 mol

Step 3: Use the stoichiometry of the balanced equation to determine the number of moles of O2 required.
According to the balanced equation, 1 mole of C6H14 reacts with (19/2) moles of O2.
Therefore, the number of moles of O2 required = 0.125 mol × (19/2) mol O2 per mol C6H14
≈ 0.594 mol O2

Step 4: Calculate the mass of oxygen required
Mass = Number of moles × Molar mass
Mass of O2 required = 0.594 mol × 32.00 g/mol
≈ 19 g

Therefore, approximately 19 grams of oxygen (O2) is required to react completely with 10.8 grams of C6H14.

To determine the mass of oxygen (O2) required to react completely with 10.8 g of C6H14, we need to use stoichiometry. Stoichiometry is a method that allows us to determine the relationship between reactants and products in a chemical equation.

Let's start by balancing the equation:
C6H14 + O2 → CO2 + H2O

The balanced equation can be simplified to:
1 C6H14 + 19/2 O2 → 6 CO2 + 7 H2O

Now, let's calculate the molar mass of C6H14:
C: 12.01 g/mol x 6 = 72.06 g/mol
H: 1.01 g/mol x 14 = 14.14 g/mol

Total molar mass of C6H14: 72.06 g/mol + 14.14 g/mol = 86.20 g/mol

Next, we need to determine the molar mass of O2:
O: 16.00 g/mol x 2 = 32.00 g/mol

Now, we can set up a proportion to find the required mass of O2:
(10.8 g C6H14) / (86.20 g/mol C6H14) = (x g O2) / (32.00 g/mol O2)

Cross-multiplying:
10.8 g C6H14 x 32.00 g/mol O2 = 86.20 g/mol C6H14 x (x g O2)

Simplifying:
345.6 g·mol/g C6H14 = 86.20 g/mol C6H14 · x g O2

Dividing both sides by 86.20 g/mol C6H14:
345.6 g·mol/g C6H14 ÷ 86.20 g/mol C6H14 = x g O2

x ≈ 4 g O2

Therefore, 4 grams of oxygen (O2) is required to react completely with 10.8 g of C6H14.

This is a stoichiometry problem. Here is a worked example; just follow the steps.

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