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Mademuasel [1]
2 years ago
6

During an experiment, the percent yield of calcium chloride from a reaction was 82.38%. Theoretically, the expected amount shoul

d have been 105 grams. What was the actual yield from this reaction? CaCO3 + HCl → CaCl2 + CO2 + H2O
Chemistry
1 answer:
gregori [183]2 years ago
7 0

Answer:

Actual yield = 86.5g

Explanation:

Percent yield = 82.38%

Theoretical yield = 105g

Actual yield = x

Equation of reaction,

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

Percentage yield = (actual yield / theoretical yield) * 100

82.38% = actual yield / theoretical yield

82.38 / 100 = x / 105

Cross multiply and make x the subject of formula

X = (105 * 82.38) / 100

X = 86.499g

X = 86.5g

Actual yield of CaCl₂ is 86.5g

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2 years ago
An amount of solid barium chloride, 20.8 g, is dissolved in 100 g water in a coffee-cup calorimeter by the reaction: BaCl2 (s) 
mamaluj [8]

Answer : The enthalpy change during the reaction is -6.48 kJ/mole

Explanation :

First we have to calculate the heat gained by the reaction.

q=m\times c\times (T_{final}-T_{initial})

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q = heat gained = ?

m = mass of water = 100 g

c = specific heat = 4.04J/g^oC

T_{final} = final temperature = 26.6^oC

T_{initial} = initial temperature = 25.0^oC

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

q=100g\times 4.04J/g^oC\times (26.6-25.0)^oC

q=646.4J

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles barium chloride = \frac{\text{Mass of barium chloride}}{\text{Molar mass of barium chloride}}=\frac{20.8g}{208.23g/mol}=0.0998mole

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Therefore, the enthalpy change during the reaction is -6.48 kJ/mole

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2 years ago
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Answer:

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Explanation:

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