Acetylene 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.
Equation 2C2H2 + 5O2 ---> 4CO2 + 2H2O
A) the mass of oxygen used
B) the mass of carbon dioxide produced
C) the mass of water produced
My answers
A) 73.7 g
B) 162.2 g
C) 14.8 g are these corect
B is right; A and C are not. If you will post your work I will find the error and correct.
A) mass of O2 = (48.0 C2H2/26.04 g/mol)(5O2/2C2H2)(16.00 g/mol) = 73.7 g of oxygen
C) mass of H2O = (48.0 g C2H2/26.04 g/mol)(2H2O/2C2H2)(18.02 g/mol) = 14.8 g of water
Thank you for posting your work. It makes it so much easier to catch the error and it let's me know about basic problems you may have.
For A you have done everything exactly right except the last step. The molar mass of oxygen is 32 since it is O2 and not O and the correct answer is just twice what you have.
For C you're set up is ok. You must have punched a wrong buttons on your calculator.
(48.0/26.04) x (2/2) x (18.02 = about 33.
So A would be 147.5 and C) 33.22
I also have another chemistry question that i am stuck on.
a and c are right now but on c you have an answer to too many significant figures. The 48.0 limits the answer to 3 s.f. so you should round that 33.22 to 33.2 grams for mass H2O.
To solve this problem, we need to use stoichiometry, which involves using the balanced equation to relate the masses of different substances.
A) To find the mass of oxygen used, we can use the stoichiometric ratio between acetylene and oxygen in the balanced equation. According to the equation, for every 2 moles of acetylene (C2H2), we need 5 moles of oxygen (O2).
First, we need to convert the mass of acetylene to moles:
Molar mass of acetylene (C2H2) = 12.01 g/mol (C) + 2(1.01 g/mol) (H) = 26.04 g/mol
Moles of acetylene = mass of acetylene / molar mass of acetylene = 48.0 g / 26.04 g/mol = 1.842 mol
Now, using the stoichiometric ratio, we can calculate the moles of oxygen required:
Moles of oxygen = moles of acetylene * (5 moles of oxygen / 2 moles of acetylene) = 1.842 mol * (5/2) = 4.605 mol
Finally, we can convert the moles of oxygen to grams:
Mass of oxygen = moles of oxygen * molar mass of oxygen
Molar mass of oxygen (O2) = 2(16.00 g/mol) = 32.00 g/mol
Mass of oxygen = 4.605 mol * 32.00 g/mol = 147.36 g
So, the correct answer for A) is 147.36 g of oxygen used.
B) To find the mass of carbon dioxide produced, we can use the same approach as above.
Moles of carbon dioxide = moles of acetylene * (4 moles of carbon dioxide / 2 moles of acetylene) = 1.842 mol * (4/2) = 3.684 mol
Mass of carbon dioxide = moles of carbon dioxide * molar mass of carbon dioxide
Molar mass of carbon dioxide (CO2) = 12.01 g/mol (C) + 2(16.00 g/mol) (O) = 44.01 g/mol
Mass of carbon dioxide = 3.684 mol * 44.01 g/mol = 162.20 g
Therefore, the correct answer for B) is 162.20 g of carbon dioxide produced.
C) To find the mass of water produced, we can use the stoichiometric ratio between acetylene and water in the balanced equation. According to the equation, for every 2 moles of acetylene, we produce 2 moles of water.
Moles of water = moles of acetylene * (2 moles of water / 2 moles of acetylene) = 1.842 mol * (2/2) = 1.842 mol
Mass of water = moles of water * molar mass of water
Molar mass of water (H2O) = 2(1.01 g/mol) (H) + 16.00 g/mol (O) = 18.02 g/mol
Mass of water = 1.842 mol * 18.02 g/mol = 33.23 g
So, the correct answer for C) is 33.23 g of water produced.
Therefore, your answers are not correct. The correct answers are:
A) 147.36 g
B) 162.20 g
C) 33.23 g