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:
12 atoms/(6.02x10^23 atoms/mole)=1.99x10^-23 mol Fe
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.
Answer:
The salt is barium chloride.
Explanation:
Moles of barium sulfate =
According to reaction, 1 mol of barium sulfate is produced from 1 mol of 
.
Then 0.0480 moles will be produced from:
of .
Mass of used = 10.00 g
Moles of =\frac{10.00 g}{\text{Molar mass}}[/tex]
Molar mass of = 208.33 g/mol
The nearest answer to our answer is 
.
The correct answer barium chloride with molar mass of 208.2 g/mol.
Answer:
Explanation:
The theory of natural selection was explored by 19th-century naturalist Charles Darwin. Natural selection explains how genetic traits of a species may change over time. This may lead to speciation, the formation of a distinct new species.
Hello,
I believe that your answer would be <span>B. water freezes
</span>Hope this helps
