Question
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
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
Answers
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.
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.