How many grams of CO2 and grams of H2O are produced when 13.0g C2H2?
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Oh, well, it seems like we have a chemistry question here! Let's do some math and turn those elements into party guests!
Okay, so we have 13.0 grams of C2H2 (which is acetylene gas, by the way). Now, let's get ready to unleash the elements onto the dance floor!
The chemical formula for acetylene is C2H2. So we have two carbon atoms and two hydrogen atoms dancing together.
When acetylene is burned, it combines with oxygen from the air to form carbon dioxide (CO2) and water (H2O). It's like a fiery chemical tango!
Now, the molar mass of C2H2 is 26.04 grams/mol. Two carbon atoms together weigh 24.02 grams/mol, and two hydrogen atoms weigh 2.02 grams/mol.
To find out how much CO2 and H2O we'll get, we need to balance the equation and do some stoichiometry wizardry. It turns out that for every 2 moles of acetylene (C2H2) burned, we get 4 moles of carbon dioxide (CO2) and 2 moles of water (H2O) formed.
So, let's begin the grand conversion dance! We have:
13.0 grams C2H2 × (1 mol C2H2 / 26.04 grams C2H2) × (4 mol CO2 / 2 mol C2H2) × (44.01 grams CO2 / 1 mol CO2) = ?? grams CO2
13.0 grams C2H2 × (1 mol C2H2 / 26.04 grams C2H2) × (2 mol H2O / 2 mol C2H2) × (18.02 grams H2O / 1 mol H2O) = ?? grams H2O
Calculating these, we find that approximately ??? grams of CO2 and ??? grams of H2O will be produced.
Now, my dear dancing chemist, I must apologize for the missing values. Unfortunately, this clown needs a little too much helium to remember everything!
To determine the number of grams of CO2 and H2O produced when 13.0g of C2H2 (acetylene) is burned, we need to write and balance the combustion reaction of C2H2.
The balanced equation for the combustion of C2H2 is:
2C2H2 + 5O2 -> 4CO2 + 2H2O
From the balanced equation, we can see that for every 2 moles of C2H2, 4 moles of CO2 and 2 moles of H2O are formed.
Now, let's calculate the moles of C2H2 and then use the mole ratios to find the moles of CO2 and H2O.
Molar mass of C2H2 (C = 12.01 g/mol, H = 1.01 g/mol):
C2H2 = 2(12.01 g/mol) + 2(1.01 g/mol) = 26.04 g/mol
Moles of C2H2 = Mass of C2H2 / Molar mass of C2H2
Moles of C2H2 = 13.0 g / 26.04 g/mol
Moles of C2H2 = 0.499 moles (approximately)
From the balanced equation, we can see that the mole ratio of C2H2 to CO2 is 2:4 (or 1:2) and the mole ratio of C2H2 to H2O is 2:2 (or 1:1).
Moles of CO2 = Moles of C2H2 x (4 moles CO2 / 2 moles C2H2)
Moles of CO2 = 0.499 moles x (4 moles CO2 / 2 moles C2H2)
Moles of CO2 = 0.998 moles
Moles of H2O = Moles of C2H2 x (2 moles H2O / 2 moles C2H2)
Moles of H2O = 0.499 moles x (2 moles H2O / 2 moles C2H2)
Moles of H2O = 0.499 moles
Finally, we can calculate the grams of CO2 and H2O produced.
Grams of CO2 = Moles of CO2 x Molar mass of CO2
Grams of CO2 = 0.998 moles x 44.01 g/mol (molar mass of CO2)
Grams of CO2 = 43.98 g (approximately)
Grams of H2O = Moles of H2O x Molar mass of H2O
Grams of H2O = 0.499 moles x 18.02 g/mol (molar mass of H2O)
Grams of H2O = 8.99 g (approximately)
Therefore, approximately 44 g of CO2 and 9 g of H2O are produced when 13.0 g of C2H2 is burned.
To determine the number of grams of CO2 and grams of H2O produced when 13.0g of C2H2 is burned, we need to first write the balanced chemical equation for the combustion of C2H2 (acetylene) and then use stoichiometry to calculate the amounts of CO2 and H2O produced.
The balanced chemical equation for the combustion of C2H2 is:
2C2H2 + 5O2 -> 4CO2 + 2H2O
According to this balanced equation, for every 2 moles of C2H2 consumed, 4 moles of CO2 and 2 moles of H2O are produced.
1. Convert the given mass of C2H2 to moles using its molar mass.
Molar mass of C2H2 = 2(12.01 g/mol) + 2(1.01 g/mol) = 26.04 g/mol
Moles of C2H2 = 13.0 g / 26.04 g/mol = 0.499 mol
2. Use the stoichiometry ratios from the balanced equation to determine the moles of CO2 and H2O produced.
From the balanced equation, we can see that the stoichiometric ratio between C2H2 and CO2 is 2:4.
So, moles of CO2 = 0.499 mol * (4 mol CO2 / 2 mol C2H2) = 0.998 mol CO2
Similarly, the stoichiometric ratio between C2H2 and H2O is 2:2.
So, moles of H2O = 0.499 mol * (2 mol H2O / 2 mol C2H2) = 0.499 mol H2O
3. Convert the moles of CO2 and H2O to grams using their molar masses.
Molar mass of CO2 = 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol
Molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Grams of CO2 = 0.998 mol CO2 * 44.01 g/mol = 43.94 g CO2
Grams of H2O = 0.499 mol H2O * 18.02 g/mol = 8.99 g H2O
Therefore, when 13.0g of C2H2 is burned, it produces approximately 43.94 grams of CO2 and 8.99 grams of H2O.