How many moles of solute are in a 15 M solution that has a volume of 2 liters?

Chemistry

1 answer:

Gelneren [198K]3 years ago

6 0

Answer:

30 moles of solute

Explanation:

15 M = moles/2 liters

moles = Molarity * Liters

moles = 15 M * 2 Liters

moles = 30

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Assume that the maximum number of ATPs is produced (38). At pH 7, and in the presence of excess Mg2 , how much of the energy in

Lubov Fominskaja [6]

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Explanation:

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3 years ago

A solution contains an unknown amount of dissolved magnesium. Addition of

Scrat [10]

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

<h3>Mass of magnesium dissolved</h3>

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?

$mass of magnesium=\frac{0.0877 moles of Na_{2}C O_{3}x24.3 grams of magnesium }{1 mole of Na_{2}C O_{3}}$