To entertain children between the ages of 2 and 90. I enjoy popping corks from b
ID: 929383 • Letter: T
Question
To entertain children between the ages of 2 and 90. I enjoy popping corks from bottles containing vinegar and baking soda. I pour about 50 mL of vinegar into a 500-mL plastic bottle. Then I wrap about 5 g of baking soda (which is sodium bicarbonate. NaHCO_3) in one layer of tissue and drop the tissue into the bottle. 1 place a cork tightly in the mouth of the bottle and step back. The chemical reaction generates CO_2(g) that pressurizes the bottle and eventually bursts the cork into the air. Everyone smiles. CH_3CO_2H + NaHCO_3 rightarrow CH_3CO_2 + Na^+ + CO_2(g) + H_2O Acetic acid in vinegar sodium bicarbonate in baking soda Find the formula mass of acetic acid and of sodium bicarbonate. How many grams of acetic acid are required to react with 5 g of NaHCO_3? Vinegar contains -5 wt% acetic acid. How many grams of vinegar are required to react with 5 g of NaHCO_3? The density of vinegar is close to 1.0 g/mL. How many mL of vinegar are required to react with 5 g of NaHCO_3? Which is the limiting reagent when you mix 50 mL of vinegar with 5 g of NaHCO_3? Use the ideal gas law (Problem 1-18) to calculate how mans L of CO_2 are generated if P = 1 bar and T = 3(H) K. If there is 0.5 L of air space in the bottle, what pressure can be generated to pop the cork?Explanation / Answer
Atomic masses C:12, H:1 and O= 16, Na=23
Molar mass of acetic acid(CH3COOH)= 12+3*1+12+32+1=60 g/mole
Molar mass of NaHCO3= 23+1+12+3*16= 84 g/mole.
Mass of acetic acid =5gms Moles of acetic acid= 5/84= 0.059524
As per the reaction given, 1 mole of acetic acid reacts with one mole of NaHCO3. Moles of acetic acid= 0.059524
Mass of acetic acid =moles* Molecularweight= 0.059524*60 =3.57 gms
c) Mass of acetic acid = 3.57gms
5 wt% of 100 = 3.57 gms
3.57 gms acetic acid are required for reacting with 5 gms of NaHCO3
But vinegar contains 5% acetic acid
5 gms of CH3COOH is there in 100 gms of Vinegar
3.57 gms are there in 3.57*100/5= 71.4gms of Vinegar
Volume of vinegar= mass/ density= 71.4/1 ml =71.4 ml
d) 50ml of vinegar correspond to 50gms of vinegar (since mass= Volume*density and density= 1g/cc)
Mass of acetic acid in 50 gm =50*5/100 =2.5 gms
Required= 3.27 gms
Hene limiting reactant is acetic acid
e) Moles of CO2 generated= ( moles of limiting reactant consumed)= 2.5/60= 0.041667
From PV= nRT
V= 0.041667*0.08206*300/0.9869(bar)=1.039 L
When there is 0.5 L of air, the volume becomes= 1.039-0.5 =0.5393L
Since temperature remained constant
P1V1= P2V2
P2= P1V1/V2= 1.039* 1/0.5393 bar=1.926 bar
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