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borishaifa [10]
2 years ago
10

During an experiment, the percent yield of calcium chloride from a reaction was 85.22%. Theoretically, the expected amount shoul

d have been 113 grams. What was the actual yield from this reaction? CaCO3 + HCl → CaCl2 + CO2 + H2O
96.3 grams
99.0 grams
113 grams
121 grams
Chemistry
2 answers:
devlian [24]2 years ago
7 0

The formula to find yield is

(Actual Yield)/(Theorectical Yield) x100

Just do the math.

85.22% x 113 = 96.2986

Convert it to 3 significant figures

Ans: 96.3g

jonny [76]2 years ago
6 0

Answer : The actual yield of the reaction is, 96.3 g

Explanation :  Given,

Percent yield of the reaction = 82.22 %

Theoretical yield of the reaction = 113 g

The formula used for the percent yield will be :

\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

or,

\text{Actual yield}=\frac{(\text{Percent yield}\times \text{Theoretical yield})}{100}

Now put all the given values in this formula, we get:

\text{Actual yield}=\frac{(85.22\times 113)}{100}=96.3g

Therefore, the actual yield of the reaction is, 96.3 g

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Volume of each solution : 60 ml 20% and 40 ml 45%

<h3>Further explanation</h3>

Given

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Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b
frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}?

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(b) 2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}

(c)Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}

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Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

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For the cell notation: Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}

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Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}

For Lead, half-equation is reduction:

Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}

Multiply first half-equation for 2 and second half-equation by 3:

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The balanced redox reaction with cell notation Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)} is

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

6 0
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