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:
Explanation:
The Chemical Equation for each one of these compounds would be the following
Sodium Chloride: NaCl
This compound contains 1 Sodium atom and 1 Chlorine atom
Sodium Oxide: Na₂O
This compound contains 2 Sodium atoms and 1 Oxygen atom.
Answer:
a frisbee flaying in the air
Explanation:
Kinetic energy can be defined as an energy possessed by an object or body due to its motion.
Mathematically, kinetic energy is given by the formula;
Where;
- K.E represents kinetic energy measured in Joules.
- M represents mass measured in kilograms.
- V represents velocity measured in metres per seconds square.
Hence, an example of kinetic energy at work is a frisbee flaying in the air because it would possess energy due to its motion in the air.
