Chemistry, 31.07.2021 02:50 atefah88

Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO (s) + H2O (l) â†’ Ca(OH)2 (s) In a particular experiment, a 2.00-g sample of CaO is reacted with excess water and 2.14 g of Ca(OH)2 is recovered. What is the percent yield in this experiment? a. 107 b. 1.07 c. 2.88 d. 81.1 e. 93.3

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Answer from: ayoismeisalex

The option closest to the percentage yield is option;

d. 81.1

Explanation:

The given chemical equation of the reaction is presented as follows;

CaO (s) + H₂O (l) → Ca(OH)₂

The mass of CaO in the experiment, m = 2.00 g

The volume of water with which the CaO was reacted = Excess volume of water

Number of moles = Mass/(Molar mass)

The mass of Ca(OH)₂ recovered, actual yield = 2.14 g

The molar mass of CaO = 56.0774 g/mol

The number of moles of CaO in the reaction, n₁ = 2.00 g/(56.0774 g/mol ≈ 0.036 moles

The molar mass of Ca(OH)₂ = 74.093 g/mol

The number of moles of Ca(OH)₂ in the reaction, n₂ = 2.14 g/(74.093 g/mol) ≈ 0.029 moles

From the given chemical reaction, one mole of CaO reacts with one mole of H₂O to produce one mole of Ca(OH)₂

Therefore, 0.036 moles of CaO will produce 0.036 moles of Ca(OH)₂

Mass = Number of moles × Molar mass

The mass of 0.036 moles of Ca(OH)₂ ≈ 0.036 moles × 74.093 g/mol = 2.667348 grams

∴ The theoretical yield of Ca(OH)₂ = 2.667348 grams

$Percentage \ yield = \dfrac{Actual \ yield}{Theoretical \ yield} \times 100 \%$

The percentage yield = (2.14 g)/(2.667348 grams) × 100 = 80.23%

Therefore, the option which is closest is option d. 81.1.

Answer from: Quest

answerambulatory;

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO (s) + H2...