A number that shows the size or amount of something. Usually the number is in reference to some standard measurement, such as a meter or kilogram. Here some scales are used to measure weight.
Answer:
C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
Explanation:
Let's start by counting how many molecules we have on each side.
<u>Reactants</u>
C - 5
H - 12
O - 2
<u>Products</u>
C - 1
O - 3
H - 2
We have to balance both sides. Since we have 5 carbon atoms on the reactants side and 1 carbon atom on the products side, we'll start with that. Write 5 as the coefficient for CO₂ to balance the carbons.
C₅H₁₂ + O₂ → 5CO₂ + H₂O
Now that the carbons are balanced, let's look at the hydrogens. There's 12 on the reactants side and 2 on the products side. To balance the hydrogens, we have to write 6 as the coefficient for H₂O.
C₅H₁₂ + O₂ → 5CO₂ + 6H₂O
Now the hydrogens are balanced. All that's left to do is balance the oxygens. Let's start by counting how many atoms we have on each side.
<u>Reactants</u>
C - 5
H - 12
O - 2
<u>Products</u>
C - 5
H - 12
O - 16
So, to balance the number of atoms on both sides, write 8 as the coefficient for O₂.
C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
Lastly, let's check to make sure everything is balanced.
<u>Reactants </u>
C - 5
H - 12
O - 16
<u>Products</u>
C - 5
O - 16
H - 12
Both sides are balanced. Therefore, the balanced chemical equation is: C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O.
Hope that helps.
Answer : The limiting reagent in this reaction is, and number of moles of excess reagent is, 1.69 moles
Explanation : Given,
Mass of = 500.0 g
Mass of = 450.0 g
Molar mass of = 342.15 g/mol
Molar mass of = 74.1 g/mol
First we have to calculate the moles of and .
and,
Now we have to calculate the limiting and excess reagent.
The given chemical reaction is:
From the balanced reaction we conclude that
As, 1 mole of react with 3 mole of
So, 1.461 moles of react with moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Number of moles of excess reagent = 6.073 - 4.383 = 1.69 moles
Therefore, the limiting reagent in this reaction is, and number of moles of excess reagent is, 1.69 moles
You must remember that oxidation number of hydrogen in acids is always +1, oxidation number of oxygen in oxides & acids is always -2... metals has always oxidation number on plus!
group NO3 comes from HNO3...and oxidation number of whole acid group is always on minus and equal to the amount of hydrogen atoms in this acid... so oxidation number of NO3 = -1
we have 2 NO3 groups so 2*(-1) = -2 and that is the reason why oxidation number of Fe in this formula must be +2... because sum of all elements always gives 0!
Now we could count of oxidation number for nitrogen... we write HNO3 and start counting from right to left:
3*(-2) from oxygens + 1 from hydrogen = -5
so nitrogen must have +5 oxidation number... because sum all in formula must be 0.