Acettlene gas (C2H2) is used in welding. During one job, a welder burns 48.0 g of acetylene. Using a balanced equation for the complete combustion of acetylene, calculate the following:
Balanced Equation
a) the mass of oxygen used
b) the mass if carbon dioxide produced
c) the mass of water produced
Use your answers from. A) B) and C) to show that this equation obeys the law of conservation of mass
2C2H2 + 5O2 ==> 4CO2 + 2H2O
b. mols C2H2 burned = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols C2H2 to mols CO2.
Now convert mols CO2 to g CO2. g CO2 = mols CO2 x molar mass CO2 = ?
c is done the same way.
a is done the same way.
For the last part, grams on the left = grams on the right. Check that out to see if the law of conservation of mass is true.
To answer these questions, we need to first write the balanced equation for the complete combustion of acetylene. The balanced equation for the combustion of acetylene is as follows:
C2H2 + 5O2 -> 4CO2 + 2H2O
In this equation, it shows that one molecule of acetylene (C2H2) combines with five molecules of oxygen (O2) to produce four molecules of carbon dioxide (CO2) and two molecules of water (H2O).
Now let's calculate the mass of oxygen used:
a) Mass of oxygen used:
From the balanced equation, we can see that 1 mole of acetylene (C2H2) requires 5 moles of oxygen (O2) to react completely. The molar mass of oxygen is approximately 32 g/mol.
To calculate the mass of oxygen used, we can use the molar mass and stoichiometry as follows:
48.0 g acetylene * (1 mol acetylene / 26.038 g acetylene) * (5 mol oxygen / 1 mol acetylene) * (32.00 g oxygen / 1 mol oxygen) = 294.38 g oxygen
So, the mass of oxygen used in this reaction is approximately 294.38 g.
Next, let's calculate the mass of carbon dioxide produced:
b) Mass of carbon dioxide produced:
Again, using the balanced equation, we can see that 1 mole of acetylene (C2H2) produces 4 moles of carbon dioxide (CO2). The molar mass of carbon dioxide is approximately 44 g/mol.
To calculate the mass of carbon dioxide produced, we can use the molar mass and stoichiometry as follows:
48.0 g acetylene * (1 mol acetylene / 26.038 g acetylene) * (4 mol carbon dioxide / 1 mol acetylene) * (44.01 g carbon dioxide / 1 mol carbon dioxide) = 338.08 g carbon dioxide
So, the mass of carbon dioxide produced in this reaction is approximately 338.08 g.
Finally, let's calculate the mass of water produced:
c) Mass of water produced:
According to the balanced equation, 1 mole of acetylene (C2H2) produces 2 moles of water (H2O). The molar mass of water is approximately 18 g/mol.
To calculate the mass of water produced, we can use the molar mass and stoichiometry as follows:
48.0 g acetylene * (1 mol acetylene / 26.038 g acetylene) * (2 mol water / 1 mol acetylene) * (18.02 g water / 1 mol water) = 62.34 g water
So, the mass of water produced in this reaction is approximately 62.34 g.
To show that this equation obeys the law of conservation of mass, we can sum up the masses of the reactants and the products:
Mass of reactants = Mass of acetylene + Mass of oxygen = 48.0 g + 294.38 g = 342.38 g
Mass of products = Mass of carbon dioxide + Mass of water = 338.08 g + 62.34 g = 400.42 g
We can see that the mass of the reactants (342.38 g) is equal to the mass of the products (400.42 g), confirming that the law of conservation of mass is obeyed in this equation.