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andre [41]
3 years ago
6

During an experiment, the percent yield of calcium chloride from a reaction was 80.34%. Theoretically, the expected amount shoul

d have been 115 grams. What was the actual yield from this reaction? CaCO3 + HCl → CaCl2 + CO2 + H2O
Chemistry
2 answers:
FinnZ [79.3K]3 years ago
6 0

Answer:

The actual yield from this reaction is 92.39 grams CaCl2

Explanation:

Step 1: Data given

% yield = 80.34 %

Expected amount of Calcium chloride (CaCl2) = 115 grams

Step 2: The balance equation

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

Step 3: Calculate actual yield

% yield = 80.34 % = 0.8034  = actual yield / theoretical yield

actual yield = 0.8034 * 115.0 grams

actual yield = 92.39 grams

The actual yield from this reaction is 92.39 grams CaCl2

nalin [4]3 years ago
6 0

Answer:

i can confirm that 92.39 (or 92.4 on the exam) is correct :)

Explanation:

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Answer each of the following questions with increases, decreases, or does not change.
Nitella [24]

Answer:

a) increases

b) decreases

c) does not change

d) increases

Explanation:

The vapour pressure of a liquid is dependent on;

I) the magnitude of intermolecular forces

II) the temperature of the liquid

Hence, when any of these increases, the vapour pressure increases likewise.

Similarly, the boiling point of a liquid depends on the magnitude of intermolecular forces present because as intermolecular forces increases, more energy is required to break intermolecular bonds.

Lastly, increase in surface area of a liquid does not really affect it's vapour pressure.

7 0
3 years ago
How many grams of NH3 can be produced from 2.30 mol of N2 and excess H2.
MrRa [10]
<h3>Answer:</h3>

78.34 g

<h3>Explanation:</h3>

From the question we are given;

Moles of Nitrogen gas as 2.3 moles

we are required to calculate the mass of NH₃ that may be reproduced.

<h3>Step 1: Writing the balanced equation for the reaction </h3>

The Balanced equation for the reaction is;

   N₂(g) + 3H₂(g) → 2NH₃(g)

<h3>Step 2: Calculating the number of moles of NH₃</h3>

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ × 2

                                  = 2.3 moles × 2

                                  = 4.6 moles

<h3>Step 3: Calculating the mass of ammonia produced </h3>

Mass = Moles × molar mass

Molar mass of ammonia gas = 17.031 g/mol

Therefore;

Mass = 4.6 moles × 17.031 g/mol

         = 78.3426 g

         = 78.34 g

Thus, the mass of NH₃ produced is 78.34 g

3 0
3 years ago
Which of the following is a true for subtropical jet streams​
vodka [1.7K]

Answer: how do we answer when there are no options??

Explanation:

5 0
3 years ago
A student is asked to standardize a solution of barium hydroxide. He weighs out 0.978 g potassium hydrogen phthalate (KHC8H4O4,
Sav [38]

Answer:

(A) 0.129 M

(B) 0.237 M

Explanation:

(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:

  • 2HA + Ba(OH)₂ → BaA₂ + 2H₂O

Where A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).

We <u>convert mass of phthalate to moles</u>, using its molar mass:

  • 0.978 g ÷ 156 g/mol = 9.27x10⁻³ mol = 9.27 mmol

Now we <u>convert mmol of HA to mmol of Ba(OH)₂</u>:

  • 9.27 mmol HA * \frac{1mmolBa(OH)_{2}}{2mmolHA} = 6.64 mmol Ba(OH)₂

Finally we calculate the molarity of the Ba(OH)₂ solution:

  • 6.64 mmol / 35.8 mL = 0.129 M

(B) The reaction between Ba(OH)₂ and HCl is:

  • 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

So<u> the moles of HCl that reacted </u>are:

  • 17.1 mL * 0.129 M * \frac{2mmolHCl}{1mmolBa(OH)_2} = 4.41 mmol HCl

And the <u>molarity of the HCl solution is</u>:

  • 4.41 mmol / 18.6 mL = 0.237 M

3 0
3 years ago
What do you need to adjust to balance a chemical equation
Nataliya [291]
You always adjust numbers and it will be before the element
so you Neva add or change a subscript
8 0
3 years ago
Read 2 more answers
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