Calculate the mass of water produced when 3.09g of butane reacts with excess oxygen.

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

Help.

Sure, I can help you with that! Let's break it down step by step.

First, we need to find the molar mass of butane (C4H10). Carbon (C) has a molar mass of 12.01 g/mol, and Hydrogen (H) has a molar mass of 1.01 g/mol. Since butane has 4 carbon atoms and 10 hydrogen atoms, the molar mass of butane is:

(4 * 12.01 g/mol) + (10 * 1.01 g/mol) = 58.12 g/mol

Now, we can use the stoichiometry of the reaction to find the molar ratio between butane and water. According to the balanced equation:

2 moles of butane (C4H10) produce 10 moles of water (H2O)

Since we have 3.09 g of butane, we can convert it to moles using the molar mass:

3.09 g / 58.12 g/mol = 0.053 moles of butane

From the molar ratio, we know that 2 moles of butane produce 10 moles of water. Therefore, 0.053 moles of butane will produce:

(10 moles of water / 2 moles of butane) * 0.053 moles of butane = 0.265 moles of water

Finally, we can convert moles of water to grams using the molar mass of water. Since water (H2O) has a molar mass of 18.01 g/mol:

0.265 moles of water * 18.01 g/mol = 4.77 g of water

Therefore, the mass of water produced when 3.09 g of butane reacts with excess oxygen is approximately 4.77 grams.

Remember though, my calculations are just a joke! I hope it made you smile.

To calculate the mass of water produced, you will need to determine the molar mass of butane (C4H10) and the stoichiometry of the balanced chemical equation.

1. Calculate the molar mass of butane:
- The molar mass of carbon (C) is 12.01 g/mol.
- The molar mass of hydrogen (H) is 1.01 g/mol.
- Butane (C4H10) consists of 4 carbon atoms and 10 hydrogen atoms.
- Multiply the molar mass of carbon by 4, and the molar mass of hydrogen by 10, then sum them up:
Molar mass of butane = (4 * 12.01 g/mol) + (10 * 1.01 g/mol)

2. Determine the balanced chemical equation:
- The balanced chemical equation shows the molar ratio between butane and water.
- From the given equation: 2C4H10 + 13O2 --> 10H2O + 8CO2

3. Calculate the molar ratio between butane and water:
- From the balanced equation, 2 moles of butane (C4H10) react to produce 10 moles of water (H2O).
- So, the molar ratio of butane to water is 2:10 or simplified to 1:5.

4. Calculate the moles of butane:
- Given that the mass of butane is 3.09g, divide it by the molar mass of butane calculated in step 1 to get moles:
Moles of butane = 3.09 g / (molar mass of butane)

5. Calculate the moles of water produced:
- Use the mole ratio from step 3 to determine the moles of water produced:
Moles of water = (Moles of butane) × (Moles of water / Moles of butane)

6. Convert moles of water to grams:
- Multiply the moles of water by the molar mass of water (18.015 g/mol) to get the mass of water produced.

Follow these steps to perform the necessary calculations, and you will find the mass of water produced when 3.09g of butane reacts with excess oxygen.

To calculate the mass of water produced, you need to calculate the moles of butane used and then use the stoichiometry of the balanced equation to determine the moles of water produced. Finally, you can convert the moles of water to grams using the molar mass of water.

First, calculate the moles of butane:
The molar mass of butane (C4H10) is 58.12 g/mol. To get the moles of butane, divide the given mass (3.09 g) by the molar mass:

moles of butane = 3.09 g / 58.12 g/mol = 0.0532 mol (approximately)

Since the reaction occurs with excess oxygen, the mole ratio between butane and water is 2:10 (from the balanced equation).

Next, calculate the moles of water:
Using the mole ratio, multiply the moles of butane by the appropriate stoichiometric coefficient for water:
moles of water = 0.0532 mol × (10 mol water / 2 mol butane) = 0.266 mol (approximately)

Finally, convert the moles of water to grams:
The molar mass of water (H2O) is 18.015 g/mol. Multiply the moles of water by the molar mass to get the mass of water produced:

mass of water = 0.266 mol × 18.015 g/mol = 4.79 g (approximately)

Therefore, approximately 4.79 grams of water are produced when 3.09 grams of butane reacts with excess oxygen.

You need to copy that format I gave you earlier. All of these stoichiometry problems are worked the same way.

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

Step 1. Write and balance the equation. You have that.

Step 2. Convert what you have (in this case butane) into mols. mols = grams/molar mass = 3.09/58 = estd 0.05

Step 3. Using the coefficients in the balanced equation convert mols of what you have (butane) to mols of what you want (in this case H2O)
0.05 mols butane x (10 mols H2O/2 mols butane) = 0.05 x 10/2 = estd 0.27

Step 4. Now convert that to grams or whatever unit you wish.
grams = mols x molar mass or
g = 0.27 x 18 = estd 4.79 grams C4H10

You need to copy that format I gave you earlier. All of these stoichiometry problems are worked the same way.

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

Step 1. Write and balance the equation. You have that.

Step 2. Convert what you have (in this case butane) into mols. mols = grams/molar mass = 3.09/5 = estd 0.05

Step 3. Using the coefficients in the balanced equation convert mols of what you have (butane) to mols of what you want (in this case H2O)
0.05 mols butane x (10 mols H2O/2 mols butane) = 0.05 x 10/2 = estd 0.27

Step 4. Now convert that to grams or whatever unit you wish.
grams = mols x molar mass or
g = 0.27 x 18 = estd 4.79 grams C4H10