23.780 g of acetaldehyde (CH3CHO) reacts completely with oxygen to produce acetic acid (HC2H3O2) using manganese (II) acetate catalyst according to the equation shown below:
2CH3CHO + O2 → 2HC2H3O2
What mass of acetic acid (HC2H3O2) was produced? (Atomic mass: C-12.011 g/mol; 0-15.999 g/mol; H-1.008 g/mol).
32.428 g
64.736 g
64.856 g
32.368 g
The balanced equation tells us that 2 moles of acetaldehyde react with 1 mole of oxygen to produce 2 moles of acetic acid.
First, we need to convert the mass of acetaldehyde given to moles:
23.780 g CH3CHO * (1 mol CH3CHO/44.053 g CH3CHO) = 0.54 mol CH3CHO
Next, we use the mole ratio from the equation to determine the moles of acetic acid produced:
0.54 mol CH3CHO * (2 mol HC2H3O2/2 mol CH3CHO) = 0.54 mol HC2H3O2
Finally, we convert the moles of acetic acid produced to grams:
0.54 mol HC2H3O2 * (60.052 g HC2H3O2/1 mol HC2H3O2) = 32.414 g HC2H3O2
Rounding to three significant figures, the mass of acetic acid produced is 32.4 g. Therefore, the correct answer is 32.368 g.
To find the mass of acetic acid (HC2H3O2) produced, we first need to determine the number of moles of acetaldehyde (CH3CHO) present in the given mass. Then, using the stoichiometric coefficients in the balanced equation, we can determine the number of moles of acetic acid produced. Finally, we can convert the moles of acetic acid to grams by multiplying by its molar mass.
First, let's calculate the number of moles of acetaldehyde:
Molar mass of CH3CHO:
1 carbon atom = 12.011 g/mol
4 hydrogen atoms = 4 x 1.008 g/mol
1 oxygen atom = 15.999 g/mol
Total molar mass = 12.011 + (4 x 1.008) + 15.999 = 44.053 g/mol
Number of moles of CH3CHO = mass / molar mass = 23.780 g / 44.053 g/mol ≈ 0.539 mol
According to the balanced equation, 2 moles of CH3CHO react to produce 2 moles of HC2H3O2.
Therefore, the number of moles of HC2H3O2 produced is also 0.539 mol.
Next, let's calculate the mass of acetic acid produced:
Molar mass of HC2H3O2:
2 carbon atoms = 2 x 12.011 g/mol
4 hydrogen atoms = 4 x 1.008 g/mol
2 oxygen atoms = 2 x 15.999 g/mol
Total molar mass = (2 x 12.011) + (4 x 1.008) + (2 x 15.999) = 60.052 g/mol
Mass of HC2H3O2 = number of moles x molar mass = 0.539 mol x 60.052 g/mol ≈ 32.368 g
Therefore, the mass of acetic acid (HC2H3O2) produced is approximately 32.368 g.
Hence, the correct answer is 32.368 g.
To find the mass of acetic acid (HC2H3O2) produced, we need to use stoichiometry and the given molar masses of the elements.
First, let's calculate the molar mass of acetaldehyde (CH3CHO):
Molar mass of acetaldehyde (CH3CHO) = (3 x molar mass of carbon) + (6 x molar mass of hydrogen) + molar mass of oxygen
= (3 x 12.011 g/mol) + (6 x 1.008 g/mol) + 15.999 g/mol
= 44.05 g/mol
Next, using the balanced chemical equation provided, we can determine the stoichiometric ratio between acetaldehyde and acetic acid. According to the equation, 2 moles of acetaldehyde react to produce 2 moles of acetic acid.
Now let's calculate the number of moles of acetaldehyde:
moles of acetaldehyde = mass of acetaldehyde / molar mass of acetaldehyde
= 23.780 g / 44.05 g/mol
= 0.5396 mol
Since the stoichiometric ratio is 1:1 for acetic acid and acetaldehyde, the number of moles of acetic acid produced is also 0.5396 mol.
Finally, let's find the mass of acetic acid produced:
mass of acetic acid = moles of acetic acid x molar mass of acetic acid
= 0.5396 mol x (2 x (12.011 g/mol) + 2 x (1.008 g/mol) + 15.999 g/mol)
= 64.856 g
Therefore, the mass of acetic acid (HC2H3O2) produced is approximately 64.856 g. So, the correct answer is 64.856 g.