Given the following unbalanced equation

C4H10 (g) + O2 (g) → CO2 (g) + H2O (l)

What mass of water forms if 6.5 L of C4
H10 (g) measured at STP is allowed to react with 32.0 L of O2 (g)measured at 27°C and 1.68 atm pressure?

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

mols C4H10 = 65L /22.4L = ?
mols O2 = use PV = nRT and solve for n.

Convert mols C4H1O to mols H2O using the coefficients in the balanced equation.
Do the same to convert mols O2 to mols H2O.
Probably the answers for mols H2O will not agree which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
The g H2O = mols H2O x molar mass H2O

To solve this problem, we need to use the given stoichiometry of the chemical equation to determine the number of moles of C4H10 and O2 that react and the number of moles of H2O produced. Then, we can convert the number of moles of H2O to its mass using the molar mass of water.

Step 1: Convert the volumes of C4H10 and O2 to moles
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 L. We can use this information to convert the volumes of C4H10 and O2 to moles.

6.5 L of C4H10 = 6.5 mol of C4H10 (since 6.5 L / 22.4 L/mol = 0.2902 mol)
32.0 L of O2 at 27°C and 1.68 atm is not at STP, so we need to use the ideal gas law to convert the volume to moles. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

Convert the temperature from Celsius to Kelvin:
T (K) = T (°C) + 273.15
T (K) = 27 + 273.15 = 300.15 K

Now we can use the ideal gas law to calculate the number of moles of O2:
PV = nRT
n = PV / RT
n = (1.68 atm) * (32.0 L) / (0.0821 L * atm/(mol * K) * 300.15 K)
n ≈ 2.34 mol of O2

Step 2: Calculate the number of moles of H2O produced
By examining the balanced chemical equation, we can determine the stoichiometric relationship between C4H10 and H2O. From the equation, we can see that 1 mole of C4H10 reacts to produce 2 moles of H2O.

So, using the balanced equation, we can determine the number of moles of H2O produced:

1 mol of C4H10 → 2 mol of H2O

Since we have 6.5 mol of C4H10, we can use this stoichiometric relationship to determine the number of moles of H2O produced by multiplying the moles of C4H10 by the stoichiometric ratio.

6.5 mol of C4H10 * (2 mol of H2O / 1 mol of C4H10) = 13 mol of H2O

Step 3: Convert the number of moles of H2O to its mass
Finally, we can convert the number of moles of H2O to its mass using the molar mass of water (H2O).

The molar mass of H2O is calculated as:
2(1.01 g/mol for hydrogen) + 16.00 g/mol for oxygen = 18.02 g/mol

So, we can calculate the mass of water produced:
Mass = number of moles * molar mass
Mass = 13 mol of H2O * 18.02 g/mol ≈ 234.26 g

Therefore, approximately 234.26 grams of water will form if 6.5 L of C4H10 is allowed to react with 32.0 L of O2 at the given conditions.

To solve this problem, we need to use the stoichiometry of the balanced equation to determine the amount of water formed.

Step 1: Convert the given volumes of C4H10 and O2 to moles.

Using the Ideal Gas Law equation, we can convert the given volumes of both gases to moles.

For C4H10:
1 mole of any ideal gas at standard temperature and pressure (STP) occupies 22.4 L.
Therefore, the moles of C4H10 can be calculated as follows:
6.5 L C4H10 * (1 mol C4H10 / 22.4 L C4H10) = 0.29 mol C4H10

For O2:
The given volume of O2 is measured at 27°C and 1.68 atm pressure, not at STP. Therefore, we need to apply the ideal gas equation to account for the non-STP conditions.

PV = nRT

P: pressure = 1.68 atm
V: volume = 32.0 L
n: moles (unknown)
R: ideal gas constant = 0.0821 L·atm/(mol·K)
T: temperature = 27 + 273.15 (convert to Kelvin) = 300.15 K

Solving for moles (n):
(1.68 atm)(32.0 L) = n (0.0821 L·atm/(mol·K))(300.15 K)
n ≈ 2.10 mol O2

Step 2: Use the stoichiometry of the balanced equation to determine the moles of water formed.

From the given equation, the stoichiometric ratio between C4H10 and H2O is 1:5. This means that for every mole of C4H10, 5 moles of H2O are produced.

Using this stoichiometric ratio, we can calculate the moles of H2O formed:

0.29 mol C4H10 * (5 mol H2O / 1 mol C4H10) = 1.45 mol H2O

Step 3: Convert the moles of water to grams.

To convert moles of H2O to grams, we need to use the molar mass of water, which is 18.015 g/mol.

Using this molar mass, we can calculate the mass of H2O:

1.45 mol H2O * (18.015 g / 1 mol H2O) ≈ 26.12 g H2O

Therefore, approximately 26.12 grams of water will form in this reaction.