A reaction that produces 14.2 grams of a product and the theoretical yield of that product is 17.1 grams is true for the following statements :
The percent yield of the product is 83.0%
The actual yield of the product is 14.2 grams.
<h3>Percentage Yield:</h3>
Percent yield is the percent ratio of actual yield to the theoretical yield.
Mathematically,
percent yield = actual yield / theoretical yield x 100%
actual yield = 14.2 grams
theoretical yield = 17.1 grams
percentage yield = 14.2 / 17.1 × 100%
percentage yield = 83.0409356725 %
percentage yield = 83.0 %
Therefore,
The percent yield of the product is 83.0%
The actual yield of the product is 14.2 grams.
learn more on percentage yield here; brainly.com/question/4180677
Answer:
ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given) (exothermic)
Explanation:
3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g) + 17H₂O(g) + 10CO₂(g)
ΔH⁰(f): 3(-365.6)Kj 1(-301)Kj 14(0)Kj 3(0)Kj 17(-241.8)Kj 10(-393.5)Kj
= -1096.8Kj = -301Kj = 0Kj = 0Kj = -4110.6Kj = -3930.5Kj
ΔHₙ°(rxn) = ∑
(ΔH˚(f)products) - ∑(ΔH˚(f)reactants)
= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]
= [-(8041.1) - (-1397.8)]Kj
= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)
∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)
ΔH°(rxn for 14.11g NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given)
The answer is the solution saturated than more dissolved with is C
Answer:
I don't know
Explanation:
Maybe they shouldn't copy each other
