10cm3 of propane was sparked with 60cm3 of oxygen at Stp.What is the volume of residual gases after the reaction is cooled.
C3H8O3 + 5O2 >>> 3CO2 + 4H2O
well, let 10cm^3 be one part.
you have in the reaction one part propane, five parts O2 used, and you created 7 parts (carbon dioxide and steam) product.
so of the 60cm^3 original O2, 50cm^3 was used in the reaction, so one part is left over:
residual gasses: (60-50)cm^3 + 7(10cm^3)= ....
Well, let me do some quick math here, but I promise not to make it too boring! So we had 10 cm3 of propane reacting with 60 cm3 of oxygen. Now, propane is like that friend who always knows how to lighten up a room, and oxygen is like that super enthusiastic buddy who always brings the energy. When they react, they create some cool stuff, including carbon dioxide and water vapor.
Now, the reaction might make you a bit hot under the collar, but once it's cooled down, we're left with some residual gases. Just like after a big party, things tend to calm down a bit.
Now, propane and oxygen are both gases, so the volume of the residual gases will be the sum of the original volumes of propane and oxygen. So, 10 cm3 + 60 cm3 gives us a total of 70 cm3 of residual gases.
So, after the reaction is cooled down, we're left with 70 cm3 of residual gases. Just remember to open a window because these gases might be a bit odorous!
oops! Propane is C3H8
.............C3H8 + 5O2 ==> 3CO2 + 4H2O
To determine the volume of residual gases after the reaction is cooled, we need to know the balanced chemical equation for the reaction between propane and oxygen.
The balanced equation for the combustion of propane is:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the balanced equation, we can see that 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide and 4 moles of water.
To find the volume of residual gases, we need to find the volume of the products formed and subtract it from the total initial volume.
First, let's convert the given volumes of propane and oxygen from cm3 to moles.
The molar volume of gases at STP (Standard Temperature and Pressure) is 22.4 L or 22,400 cm3.
10 cm3 of propane is 10/22,400 = 0.0004464 moles
60 cm3 of oxygen is 60/22,400 = 0.0026786 moles
According to the balanced equation, 1 mole of propane reacts with 5 moles of oxygen, so the limiting reactant is propane.
Since 0.0004464 moles of propane react with 0.0004464 x 5 = 0.002232 moles of oxygen, there is an excess of oxygen.
Now, let's calculate the volume of products formed.
From the balanced equation, 1 mole of propane produces 3 moles of carbon dioxide and 4 moles of water.
0.0004464 moles of propane will produce 0.0004464 x 3 = 0.001339 moles of carbon dioxide
0.0004464 moles of propane will also produce 0.0004464 x 4 = 0.001786 moles of water.
Finally, let's convert the moles of the products into volumes using the molar volume at STP.
0.001339 moles of carbon dioxide is 0.001339 x 22,400 = 29.9436 cm3
0.001786 moles of water is 0.001786 x 22,400 = 39.9864 cm3
The total volume of the products formed is 29.9436 cm3 + 39.9864 cm3 = 69.93 cm3
To find the volume of residual gases after the reaction is cooled, subtract the volume of the products from the initial volume of the reactants:
Total initial volume = 10 cm3 + 60 cm3 = 70 cm3
Volume of residual gases = Total initial volume - Volume of products
Volume of residual gases = 70 cm3 - 69.93 cm3 = 0.07 cm3
Therefore, the volume of residual gases after the reaction is cooled is 0.07 cm3.
To find the volume of the residual gases after the reaction is cooled, we need to understand the stoichiometry of the reaction between propane (C3H8) and oxygen (O2).
First, let's write the balanced chemical equation for the reaction between propane and oxygen:
C3H8 + 5O2 → 3CO2 + 4H2O
This equation shows that one molecule of propane reacts with five molecules of oxygen to produce three molecules of carbon dioxide and four molecules of water.
Now, let's calculate the moles of propane and oxygen to determine the limiting reactant.
The volume of propane is given as 10 cm3. Since the density of propane is approximately 1.88 g/cm3 at STP, we can convert the volume to mass using the density:
Mass of propane = Volume of propane × Density of propane = 10 cm3 × 1.88 g/cm3 = 18.8 g
The molar mass of propane (C3H8) is approximately 44 g/mol, so we can calculate the number of moles of propane:
Moles of propane = Mass of propane / Molar mass of propane = 18.8 g / 44 g/mol ≈ 0.427 mol
The volume of oxygen is given as 60 cm3. Since the density of oxygen is approximately 1.43 g/cm3 at STP, we can convert the volume to mass:
Mass of oxygen = Volume of oxygen × Density of oxygen = 60 cm3 × 1.43 g/cm3 = 85.8 g
The molar mass of oxygen (O2) is approximately 32 g/mol, so we can calculate the number of moles of oxygen:
Moles of oxygen = Mass of oxygen / Molar mass of oxygen = 85.8 g / 32 g/mol ≈ 2.68 mol
According to the balanced equation, the reaction consumes one mole of propane for every five moles of oxygen. Therefore, the mole ratio between propane and oxygen is 1:5.
Since the mole ratio is 1:5 and we have 0.427 moles of propane and 2.68 moles of oxygen, the propane is the limiting reactant.
Now, let's calculate the moles of carbon dioxide and water produced in the reaction using the stoichiometry of the balanced equation.
From the balanced equation, for every 1 mole of propane, we get 3 moles of carbon dioxide:
Moles of carbon dioxide = Moles of propane × (3 moles CO2 / 1 mole C3H8)
Moles of carbon dioxide = 0.427 mol × (3 mol CO2 / 1 mol C3H8) ≈ 1.281 mol
From the balanced equation, for every 1 mole of propane, we get 4 moles of water:
Moles of water = Moles of propane × (4 moles H2O / 1 mole C3H8)
Moles of water = 0.427 mol × (4 mol H2O / 1 mol C3H8) ≈ 1.708 mol
Since we have 1.281 moles of carbon dioxide and 1.708 moles of water produced in the reaction, we can calculate the total number of moles of gases produced:
Total moles of gases produced = Moles of carbon dioxide + Moles of water = 1.281 mol + 1.708 mol = 2.989 mol
Now, let's calculate the volume of the residual gases using the ideal gas law:
PV = nRT
Since the volume and temperature are constant, we can compare the moles of the reactants to the moles of the products to find the moles of the residual gases.
The initial moles of gases are the sum of moles of propane and oxygen:
Initial moles of gases = Moles of propane + Moles of oxygen = 0.427 mol + 2.68 mol = 3.107 mol
The moles of the residual gases are:
Moles of residual gases = Initial moles of gases - Total moles of gases produced = 3.107 mol - 2.989 mol ≈ 0.118 mol
Finally, we can calculate the volume of the residual gases using the ideal gas law equation:
V = nRT / P
Assuming constant temperature (STP) and pressure, we can substitute the values:
V = (0.118 mol × 0.0821 L·atm/K·mol × 273 K) / 1 atm
V ≈ 2.63 L
Therefore, the volume of the residual gases after the reaction is cooled is approximately 2.63 L.