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Effectus [21]
3 years ago
7

During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.

mL of the base was required to neutralize the sample. How many grams of HCl are present in 500. mL of this acid solution?
Chemistry
1 answer:
Travka [436]3 years ago
4 0

<u>Answer:</u> The mass of HCl present in 500 mL of acid solution is 36.5 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL

Putting values in above equation, we get:

1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

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The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.

Further Explanation:

To solve this problem, follow the steps below:

  1. Write the balanced chemical equation for the given reaction.
  2. Convert the mass of calcium carbonate into moles.
  3. Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
  4. Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}

STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.

mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}

STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.

For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,

mol \ CaO \ = 0.2498 \ mol

STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.

mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g

Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.

Therefore,

\boxed {mass \ CaO \ = 14 \ g}

Learn More

  1. Learn more about stoichiometry brainly.com/question/12979299
  2. Learn more about mole conversion brainly.com/question/12972204
  3. Learn more about limiting reactants brainly.com/question/12979491

Keywords: thermal decomposition, stoichiometry

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