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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6.4 x 10^-7 = [CO]^2[O2]/ [CO2}^2 = ( 2.0 x 10^-3)^2 ( 1.0 x 10^-3)/ [CO2]^2 =
<span>=4.0 x 10^-9 / [CO2]^2 </span>
<span>[CO]= sq.rt ( 4.0 x 10^-9)/ 6.4 x 10^-7=7.9 x 10^-2 M </span>
<span>2.6 x 10^-3 = [I]^2 / [I2] = [I]^2 / 0.95 </span>
<span>[I]= sq.rt ( 2.6 x 10^-3 x 0.95)=5.0 x 10^-2 M </span>
<span>Ksp = [Ba2+][CO32-] = ( 1.1 x 10^-4)^2=1.2 x 10^-8</span>
Answer:
Explanation:
We can use the Ideal Gas Law to solve this problem
pV = nRT
Data:
p = 1.1 atm
n = 3.2 mol
R = 0.082 16 L·atm·K⁻¹mol⁻¹
T = 295 K
Calculation:
Answer:
all calcium atoms have an atomic number of 20 but there are three isotopes of calcium - 39, 40, and 42
