The answer is 8. Hope this helps.
solid
s
liquid
l
gas
g
aqueous solution
aq
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Answer:
2200 L
Explanation:
Ideal gas law:
PV = nRT,
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
The initial number of moles is:
(110 atm) (80 L) = n (0.0821 L atm / K / mol) (30 + 273.15) K
n = 353.58 mol
After some gas is removed, the number of moles remaining is:
(80 atm) (80 L) = n (0.0821 L atm / K / mol) (30 + 273.15) K
n = 257.15 mol
The amount of gas removed is therefore:
n = 353.58 mol − 257.15 mol
n = 92.43 mol
At normal conditions, the volume of this gas is:
PV = nRT
(1 atm) V = (92.43 mol) (0.0821 L atm / K / mol) (273.15 K)
V = 2162.5 L
Rounded, the volume is approximately 2200 liters.
The percent yield of CO₂ is 93.3%.
<h3>What is the percent yield of CO₂?</h3>
The percent yield of a substance is given as follows:
- Percent yield = actual yield/theoretical yield * 100 %
The equation of the reaction is used to determine the theoretical yield.
- NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂
Moe ratio of sodium bicarbonate and CO₂ is 1 : 1.
Given that sodium bicarbonate is the limiting reactant, the theoretical yield of CO₂ will be:
Moles of NaHCO₃ reacting = 2.01/84 = 0.0239 moles
Theoretical yield of CO₂ = 0.0239 moles * 22.4L/mol = 0.536 L
Actual yield = 0.50 L
Percent yield = 0.50/0.536 * 100%
Percent yield = 93.3%
In conclusion, the percent yield is the ratio of the actual yield and theoretical yield.
<em>Note that the complete question is given below:</em>
<em>Calculate your % yield of co2 in the reaction based on the grams of nahco3 being the limiting reagent in the reaction between 2.01 g of sodium bicarbonate and 24.6 mL of 1.5 M acetic acid? They produce 0.50 L of at s.t.p.</em>
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Learn mores about percent yield at: brainly.com/question/8638404
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