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tester [92]
3 years ago
6

In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what

is the percent yield?
Chemistry
1 answer:
melomori [17]3 years ago
8 0

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

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<span>Answer: Correct answers are- Electron affinity decreases; Cl has 7 valence electrons but Na has only 1. So Na is going to lose its e, Cl is going to gain an e.</span>
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3 years ago
2. In Experiment SOL, you investigated the solubility of oxalic acid. Sodium oxalate, Na2C2O¬4, is the sodium salt of this acid.
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Answer:

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The oxalic acid in water has two dissociations.

Explanation:

Na2C2O4 ---> 2Na+   +  C2O4-2

Sodium oxalate is the conjugate base of a weak acid. In water this salt, dissociates completely giving rise to the sodium and oxalate ions. As Na+ comes from a strong base, in water it does not produce hydrolysis while oxalate does react in water, because it takes a proton from it and it generates a basic hydrolysis releasing OH-.

C2O4-2  + H2O ⇄  HC2O4-  +  OH-

In water the salt is basic.  The pH of an aqueous solution of this salt is basic, since OH- is generated.

The HC2O4- has a second hydrolisis, it takes another proton from water to form oxalic acid.

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The oxalic acid acts as a weak acid, it can release 2 protons to water, to make oxalate (its conjugate base).

H2C2O4  + H2O ⇄ H3O+  + HC2O4-

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6 0
3 years ago
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