When C2H4O2 burns in oxygen, the products are carbon dioxide and water. How many grams of water are obtained from burning 4.76 g of this substance?
How would I solve this problem?
1. Write and balance the equation.
2. Convert g C2H4O to mols. mols = grams/molar mass.
3. Using the coefficients in the balanced equation, convert mols C2H4O to mols H2O.
4. Now convert mols H2O to grams. g = mols x molar mass.
I did that and got 2.86 g H20. It was correct, thank you!
Thank you for letting me know.
To solve this problem, you will need to use stoichiometry and the molar masses of the substances involved. Stoichiometry is a branch of chemistry that deals with the relationship between the mass, moles, and balanced chemical equations.
Here's how you can solve the problem step by step:
1. Write the balanced chemical equation for the burning of C2H4O2 (acetic acid) in oxygen:
C2H4O2 + O2 → CO2 + H2O
2. Determine the molar mass of acetic acid (C2H4O2):
C: 12.01 g/mol
H: 1.01 g/mol (4 hydrogen atoms in acetic acid)
O: 16.00 g/mol (2 oxygen atoms in acetic acid)
Molar mass of acetic acid = (2 * 12.01) + (4 * 1.01) + (2 * 16.00) = 60.05 g/mol
3. Calculate the number of moles of acetic acid (C2H4O2) in 4.76 g:
Moles = Mass / Molar mass
Moles of C2H4O2 = 4.76 g / 60.05 g/mol
4. Use the balanced chemical equation to determine the ratio between the number of moles of acetic acid and water. From the equation, we can see that 1 mole of acetic acid produces 1 mole of water.
5. Convert the moles of acetic acid to moles of water. Since the ratio is 1:1, the number of moles of water will be the same.
6. Calculate the mass of water produced from the moles of water using the molar mass of water (H2O):
Molar mass of H2O = (2 * 1.01) + (1 * 16.00) = 18.02 g/mol
Mass of water = Moles of water * Molar mass of H2O
7. Substitute the moles calculated in step 4 into the equation from step 6 to find the mass of water obtained.
Finally, solve for the mass of water by substituting the values into the equation, and you will get the answer to your question.