Question:
C25H52(s) + 38O2(g)----> 25CO2(g) + 26H2O
How many liters of carbon dioxide are made when 23.4 grams of paraffin (C25H52)burns?
I got 13,000 for my answer.
Is that right??
Thanks:)
You start with 0.0664 moles of paraffin and make 25 times as many moles of CO2, which is 1.66 moles. That is a lot less than 13,000 liters, at any reasonable pressure and temperature. You should specify what P and T they are assuming.
Thanks:) Can you please help me with my other chemistry question. I posted it down below! Thank you so much:)
See below.
To find the correct answer, we need to use stoichiometry, which involves converting the given mass of paraffin to the volume of carbon dioxide produced.
Step 1: Find the molar mass of paraffin (C25H52):
C25H52 = (25 x atomic mass of carbon) + (52 x atomic mass of hydrogen)
= (25 x 12.011 g/mol) + (52 x 1.008 g/mol)
= 300.275 g/mol + 52.416 g/mol
= 352.691 g/mol
Step 2: Convert the given mass of paraffin to moles using its molar mass:
23.4 g paraffin x (1 mol paraffin / 352.691 g) = 0.06629 mol
Step 3: Use the balanced equation to determine the moles of carbon dioxide produced:
From the balanced equation, we know that 1 mol of paraffin produces 25 mol of carbon dioxide.
Therefore, 0.06629 mol paraffin x (25 mol CO2 / 1 mol paraffin) = 1.65725 mol CO2
Step 4: Convert the moles of carbon dioxide to volume using the ideal gas law:
Using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature:
V = (nRT) / P
Assuming the temperature and pressure are constant, we can use the molar volume of a gas at standard temperature and pressure (STP), which is 22.4 L/mol.
V = (1.65725 mol CO2) x (22.4 L/mol) = 37.0584 L CO2
Therefore, the correct answer is approximately 37.06 liters of carbon dioxide.
Your answer of 13,000 liters seems incorrect.