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Taking into account the reaction stoichiometry, 2.13 grams of magnesium was dissolved in the solution.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
Mg²⁺(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2 Na⁺(aq)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Mg²⁺: 1 mole
- Na₂CO₃: 1 mole
- MgCO₃: 1 mole
- Na⁺: 2 moles
The molar mass of the compounds is:
- Mg²⁺: 24.3 g/mole
- Na₂CO₃: 106 g/mole
- MgCO₃: 84.3 g/mole
- Na⁺: 23 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Mg²⁺: 1 mole ×24.3 g/mole= 24.3 grams
- Na₂CO₃: 1 mole ×106 g/mole= 106 grams
- MgCO₃: 1 mole ×84.3 g/mole=84.3 grams
- Na⁺: 2 moles ×23 g/mole= 46 grams
The following rule of three can be applied: If by reaction stoichiometry 1 mole of Na₂CO₃ react with 24.3 grams of magnesium, 0.0877 moles of Na₂CO₃ react with how much mass of magnesium?
<u><em>mass of magnesium= 2.13 grams</em></u>
Finally, 2.13 grams of magnesium was dissolved in the solution.
Learn more about the reaction stoichiometry:
Answer:
<h2>127.57 moles</h2>Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
We have the final answer as
<h3>127.57 moles</h3>Hope this helps you
Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
- V₂ = 295 mL
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L