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2 years ago

Calculate the molarity of solution of "sodium sulfate" that contains 5.2 grams sodiums sulfate diluted to 500mL

1 answer:

3 0

Taking into account the definition of molarity, the molarity of solution of sodium sulfate is 0.0732 $\frac{moles}{liter}$ .

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$Molarity=\frac{number of moles}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

<h3>Molarity of solution of sodium sulfate.</h3>

In this case, you have:

number of moles of sodium sulfate= $5.2 grams\frac{1 mole}{142 grams} = 0.0366 moles$ (being 142 g/mole the molar mass of sodiums sulfate)
volume= 500 mL= 0.5 L (being 1000 mL= 1 L)

Replacing in the definition of molarity:

$Molarity=\frac{0.0366 moles}{0.5 L}$

Solving:

Molarity= 0.0732 $\frac{moles}{liter}$

Finally, the molarity of solution of sodium sulfate is 0.0732 $\frac{moles}{liter}$ .

Learn more about molarity:

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part of the SO2 that is introduced into the atmosphere by combustion of sulfur containing compounds ends up being converted to s

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5 mol.

Explanation:

Equation of the reaction

2SO2 + 2H2O + O2 --> 2H2SO4

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3 years ago

Lithium reacts with bromine (Br2) in a synthesis reaction to produce lithium bromide. Determine the limiting reactant if 25.0 gr

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Answer: Bromine is the limiting reactant

Explanation:

First of all let's generate a balanced equation for the reaction

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From the question given, were told that 25g of Li and 25g Br2 were present at the take-off of the reaction. Converting these Masses to mole, we have:

Number of mole of Li = 25/7 = 3.6moles

Number of mole of Br2 = 25/160 = 0.156mol.

To know which is the limiting reactant, we have to compare the ratio of the number of mole of experimental Li and Br2 to that of theoretical Li and Br2

For the experimental yield:

Li : Br2 = 3.6/ 0.156 = 23 : 1

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From the above, we see clear that Br2 is the limiting reactant because according to the equation( which gives the theoretical yield), for every 2moles of Li, 1mole of Br2 is used up. But this is not so from the experiment conducted as 23moles required 1mole of Br2.

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