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Ymorist [56]
2 years ago
6

A lab technician needs to create 570.0 milliliters of a 2.00 M solution of magnesium chloride (MgCl2). To make this solution, ho

w many grams of magnesium chloride does the technician need?
Refer to the periodic table for help. Express your answer to three significant figures.
Chemistry
1 answer:
love history [14]2 years ago
7 0

Answer:

108.3g

Explanation:

Given parameters:

Volume of solution = 570mm = 0.57L (1000mm = 1L)

Molarity of solution = 2M

Unknown

Mass of MgCl₂

Solution

We first find the  number of moles in the given concentration of magnessium chloride using the expression below:

       Number of moles of MgCl₂= Molarity x volume = 2 x 0.57 = 1.14moles

Using this known moles, we can the unknown mass of MgCl₂ the technician would require:

       Mass of MgCl₂ required = number of moles of MgCl₂ x molar mass

Molar mass of MgCl₂:

Atomic mass of Cl = 35.5g

Atomic mass of Mg = 24g

MgCl₂ = 24 + (35.5x2) = 95gmol⁻¹

Mass of MgCl₂ required = 1.14mole x 95gmole⁻¹ = 108.3g

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<h3>Explanation</h3>

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Start with the balanced chemical equation for this process:

\text{K}_2\text{CrO}_4 \; (aq) + \text{Ba}(\text{NO}_3)_2 \; (aq) \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{KNO}_3 \; (aq)

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\text{K}^{+} and \text{CrO}_4^{-} are found on both sides of the equation by the same quantity. The two ions thus took no part in the net reaction and act as spectator ions.

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Explanation:

Please, observe the solution in the attached Word document.

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