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Kisachek [45]
3 years ago
15

__MgCl2 +

Chemistry
1 answer:
Anna35 [415]3 years ago
6 0

Answer:

24.5 g of NaCl

Explanation:

We begin from the balanced reaction:

3MgCl₂ +   2Na₃PO₄  →    6NaCl   +  Mg₃(PO₄)₂

If the sodium phosphate is in excess, then the limting reagent is the magnessium chloride.

We convert mass to moles:

20 g . 1mol / 95.2g = 0.210 moles.

3 moles of MgCl₂ can produce 6 moles of NaCl

0.210 moles of salt, may produce (0.210 . 6) /3 = 0.420 moles

Ratio of reactant is twice the product

We convert the moles to mass:

0.420 mol .  58.45 g/mol = 24.5 g

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A 1.0 M solution of a compound with 2 ionizable groups (pKa's = 6.2 and 9.5; 100 mL total) has a pH of 6.8. If a biochemist adds
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Answer:

pH = 9,32

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The compound with 2 ionizable groups has the following equilibriums:

H₂M ⇄ HM⁻ + H⁺ pka = 6,2

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The reaction of M²⁻ with HCl is:

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The moles of M²⁻ are:

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And moles of HCl are:

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That means that moles of M²⁻ will be 0,1-0,06 = 0,04mol and moles of HM⁻ will be the same than HCl, 0,06mol

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [M²⁻] / [HM⁻]

Replacing:

pH = 9,5 + log₁₀ [0,04] / [0,06]

<em>pH = 9,32</em>

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I hope it helps!

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