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Ksivusya [100]
3 years ago
7

In an ionic bond, Select one:

Chemistry
1 answer:
Elena-2011 [213]3 years ago
8 0

Answer:

A

Explanation:

B - Can't be - Ionic bonds are formed by oppositely charged ions

C - Can't be - This is a covalent bond

D - Can't be - This is an induced dipole or a van Der waals attraction

E - This is a permanent dipole dipole attraction

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Fe2O3(s) + 3CO(g) ---&gt; 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast
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Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles

\text{Moles of} CO=\frac{260}{28}=9.3moles

Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)

According to stoichiometry :

1 mole of Fe_2O_3 require 3 moles of CO

Thus 2.8 moles of Fe_2O_3 will require=\frac{3}{1}\times 2.8=8.4moles  of CO

Thus Fe_2O_3 is the limiting reagent as it limits the formation of product and CO is the excess reagent.

As 1 mole of Fe_2O_3 give = 2 moles of Fe

Thus 2.8 moles of Fe_2O_3 give =\frac{2}{1}\times 2.8=5.6moles of Fe

Mass of Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%

Therefore, the percent yield is, 91.8%

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