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nikitadnepr [17]
3 years ago
15

¿Cuántos moles de C3H8 contienen 9.25 x 10^24 moléculas?

Chemistry
1 answer:
Igoryamba3 years ago
5 0

Answer:

\large \boxed{\text{e. 15.4 mol C$_{3}$H}_{8}}

Explanation:\text{Moles of propane} = 9.25 \times 10^{24}\text{ molecules} \times \dfrac{\text{1 mol }}{6.022 \times\ 10^{23} \text{molecules}} = \textbf{15.4 mol}

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You have to use Avogadro's number (6.02x10^23 molecules/mole) to find the number of moles each reactant starts off with.
moles of Fe and O₂:
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6 molecules/(6.02x10^23 molecules/mole)=9.967x10^-24 mol <span>O₂
</span>Then you find the limiting reagent by finding how much product each given amount of reactant can make.  Which ever one produces the least amount of product is the limiting reagent.
amount of Fe₂O₃ produced:
<span>(1.99x10^-23 mol Fe)x(2mol/4mol)= 9.967x10^-24mol Fe</span>₂O₃<span> 
</span>(9.967x10^-24 mol O₂)x(2mol/3mol)= 6.645x10^-24 mol Fe₂O₃<span> 
</span>since oxygen produces the leas amount of product, oxygen is the limiting reagent.  since we know that oxygen is the limiting reagent we can use the amount of product formed with oxygen to find the amount of iron used.
6.645x10^-24 mol Fe₂O₃x(4mol/2mol)=1.329x10^-23 mol Fe consumed
<span> find the amount left over by subtracting the original amount of Fe by the amount consumed in the reaction.
</span>1.993x10^-23-1.329x10^-23= 6.645x10^-23mol Fe left
 find the number of atoms by multiplying that by Avogadro's number.
<span>(6.645x10^-23mol)x(6.02x10^23 atoms/mol)=4 atoms
</span>therefore 4 atoms of Fe will be left over after the reaction happens.

I hope this helps.
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For argument’s sake, let’s say that the mass of magnesium oxide was 3 g and that of the oxide was 5 g.  

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The 3 g of magnesium must have combined with 2 g of oxygen to form 5 g of magnesium oxide.

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