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?

Would it be 26.1 (g)

13/2

To solve this problem, we need to follow a few steps:

Step 1: Convert the given volumes of gases to moles.
Step 2: Use the mole ratio from the balanced equation to determine the moles of water formed.
Step 3: Convert the moles of water to grams.

Let's break down each step:

Step 1: Convert the given volumes of gases to moles.
Given the volume of C4H10 is 6.5 L and the temperature and pressure are STP (Standard Temperature and Pressure), we can use the ideal gas law to convert the volume to moles.

Using the ideal gas law equation: PV = nRT
Rearranging the equation: n = PV/RT

P = pressure in atm (STP = 1 atm)
V = volume in liters
n = moles of gas
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature in Kelvin (STP = 273 K)

n(C4H10) = (1 atm)(6.5 L)/(0.0821 L•atm/mol•K)(273 K)
n(C4H10) = 3.06 moles

Given the volume of O2 is 32.0 L and the temperature and pressure are 27°C and 1.68 atm respectively, we need to convert these conditions to STP.

Convert temperature from °C to Kelvin:
T(K) = T(°C) + 273
T(K) = 27 + 273
T(K) = 300 K

Now, let's use the ideal gas law to convert the volume and pressure to moles at STP.

n(O2) = (1 atm)(32.0 L)/(0.0821 L•atm/mol•K)(300 K)
n(O2) = 1.52 moles

Step 2: Use the mole ratio from the balanced equation to determine the moles of water formed.
From the balanced equation, we can see that the mole ratio between C4H10 and H2O is 1:2. Therefore, for every 1 mole of C4H10, 2 moles of H2O are formed.

Based on the moles we calculated in Step 1:
n(C4H10) = 3.06 moles
Therefore, moles of H2O formed = 2 x n(C4H10) = 2 x 3.06 = 6.12 moles.

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 (H2O), which is 18.02 g/mol.

Mass(H2O) = moles(H2O) x molar mass(H2O)
Mass(H2O) = 6.12 moles x 18.02 g/mol
Mass(H2O) = 110.424 g

So, the mass of water formed is 110.424 grams.

Therefore, the correct answer is not 26.1 g.