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3. Use the balanced chemical equation from the last question to solve this situation: You combine 0.5 grams of Na2CO3 with excess CaCl2. How many grams of NaCl would you expect this reaction to produce? Show all work below. g

Respuesta :

jamesesther1234

Answer:

0.27 g

Explanation:

The reaction equation:

[tex]Na_{2} CO_{3} + CaCl_{2}[/tex] → [tex]2NaCl + CaCO_{3}[/tex]

106g of Na2CO3 - 1 mole

0.5g of Na2CO3 = 0.5 ÷ 106

= 0.0047 moles.

1 mole of NaCl - 58.5

⇒ 0.0047 moles = 0.0047 × 58.5

= 0.27g.

dsdrajlin

When 0.5 grams of Na₂CO₃ react with excess CaCl₂, 0.6 g of NaCl are formed.

We combine 0.5 grams of Na₂CO₃ with excess CaCl₂ and we want to know the mass of NaCl produced. This is a stoichiometry problem.

What is stoichiometry?

Stoichiometry refers to the relationship between the quantities of reactants and products before, during, and following chemical reactions.

First, we will write the balanced chemical equation.

Na₂CO₃ + CaCl₂ ⇒ 2 NaCl + CaCO₃

We will consider the following relationships.

  • The molar mass of Na₂CO₃ is 105.99 g/mol.
  • The molar ratio of Na₂CO₃ to NaCl is 1:2.
  • The molar mass of NaCl is 58.44 g/mol.

[tex]0.5 g Na_2CO_3 \times \frac{1molNa_2CO_3}{105.99gNa_2CO_3} \times \frac{2molNaCl}{1molNa_2CO_3} \times \frac{58.44gNaCl}{1molNaCl} = 0.6gNaCl[/tex]

When 0.5 grams of Na₂CO₃ react with excess CaCl₂, 0.6 g of NaCl are formed.

Learn more about stoichiometry here: https://brainly.com/question/9743981

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