When 13.0 grams of acetylene, C2H2, is reacted, what mass of water, H2O, is produced?
To determine the mass of water produced when 13.0 grams of acetylene (C2H2) is reacted, we need to balance the chemical equation for the reaction.
The balanced equation for the reaction between acetylene and water is as follows:
C2H2 + 2H2O -> 2CO2 + 2H2
From the balanced equation, we can see that each mole of acetylene (C2H2) reacts with 2 moles of water (H2O) to form 2 moles of carbon dioxide (CO2) and 2 moles of hydrogen gas (H2).
First, let's calculate the number of moles of acetylene:
Molar mass of C2H2 = 2 * atomic mass of carbon + 2 * atomic mass of hydrogen
= 2 * 12.01 g/mol + 2 * 1.01 g/mol
= 26.04 g/mol
Number of moles of acetylene = mass of acetylene / molar mass
= 13.0 g / 26.04 g/mol
= 0.498 mol
Since the stoichiometric ratio between acetylene and water is 1:2, the number of moles of water in the reaction is double that of acetylene.
Number of moles of water = 2 * number of moles of acetylene
= 2 * 0.498 mol
= 0.996 mol
Now, let's calculate the mass of water:
Molar mass of H2O = 2 * atomic mass of hydrogen + atomic mass of oxygen
= 2 * 1.01 g/mol + 16.00 g/mol
= 18.02 g/mol
Mass of water = number of moles of water * molar mass
= 0.996 mol * 18.02 g/mol
= 17.94 g
Therefore, when 13.0 grams of acetylene is reacted, approximately 17.94 grams of water is produced.
To determine the mass of water produced when 13.0 grams of acetylene (C2H2) is reacted, you need to follow a few steps.
1. Start by writing out the balanced chemical equation for the reaction. In this case, the reaction involves acetylene (C2H2) reacting with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O):
C2H2 + O2 -> CO2 + H2O
2. Determine the molar mass of acetylene (C2H2). To do this, you can look up the atomic masses of all the elements involved (C = 12.01 g/mol, H = 1.01 g/mol) and calculate the total molar mass of acetylene:
(2 * C) + (2 * H) = (2 * 12.01 g/mol) + (2 * 1.01 g/mol) = 26.04 g/mol
This means that one mole of acetylene has a mass of 26.04 grams.
3. Convert the given mass of acetylene (13.0 grams) to moles. Use the following formula:
moles = mass (g) / molar mass (g/mol)
moles = 13.0 g / 26.04 g/mol = 0.499 moles (rounded to three decimal places)
4. Examine the balanced equation to determine the ratio of acetylene (C2H2) to water (H2O). From the equation, you can see that 1 mole of acetylene reacts to produce 1 mole of water.
5. Use the mole ratio from the balanced equation to calculate the moles of water produced. Since the mole ratio is 1:1, the moles of water produced will also be 0.499 moles.
6. Finally, convert the moles of water produced to grams. Use the molar mass of water, which is approximately 18.02 g/mol.
mass = moles * molar mass
mass = 0.499 moles * 18.02 g/mol = 8.99 grams (rounded to two decimal places)
Therefore, when 13.0 grams of acetylene is reacted, approximately 8.99 grams of water is produced.
Reacted with what? O2?
2C2H2 + 5O2 ==> 4CO2 + 2H2O
mols C2H2 = grams/molar mass = ?
Then H2O = mols C2H2 x (2 mols H2O/2 mols C2H2) = ? mols H2O
Convert mols H2O to grams.
grams H2O = mols H2O x molar mass H2O