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kipiarov [429]
2 years ago
8

How many grams of water are produced when 35.8 grams of Calcium hydroxide reacts with lots of Hydrochloric acid?

Chemistry
1 answer:
Veseljchak [2.6K]2 years ago
5 0

Are produced 72 grams of water in this reaction.

<h3>Mole calculation</h3>

To find the value of moles of a product from the number of moles of a reactant, it is necessary to observe the stoichiometric ratio between them:

                          Ca(OH)_2 + 2HCl = > 2H_2O + CaCl_2

Analyzing the reaction, it is possible to see that the stoichiometric ratio is 1:2, so we can perform the following expression:

MM_{Ca(OH)_2} = 74.1g/mol

                                           MM = \frac{g}{mol}

                                             74.1 = \frac{35.8}{mol}\\mol = 2

So, if there are 2 mols of Ca(OH)2:

                                  Ca(OH)2    |    H2O

                                          \frac{1mol}{2mol} =\frac{2mol}{xmol}

                                             x = 4mol

Finally, just find the number of grams of water using your molar mass:

MM_{H_2O} = 18g/mol

                                              18= \frac{m}{4}\\m = 72g

So, 72 grams are produced of water in this reaction.

Learn more about mole calculation in: brainly.com/question/2845237

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A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o
Andreas93 [3]
<h3>Answer:</h3>

2.809 L of H₂SO₄

<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>
  • The reaction between NaOH and H₂SO₄ is a neutralization reaction.
  • The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

<h3>Second: We calculate the umber of moles of NaOH used </h3>
  • Number of moles = Mass ÷ Molar mass
  • Molar mass of NaOH is 40.0 g/mol
  • Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

                          = 0.33 moles

<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
  • From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
  • Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
  • Thus, Moles of H₂SO₄ = moles of NaOH × 2

                                    = 0.33 moles × 2

                                   = 0.66 moles of H₂SO₄

<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
  • When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
  • That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

                                = 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

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Black_prince [1.1K]

Answer:

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