Answer:
C. the relative number of atoms of each element, using the lowest whole ratio.
Explanation:
The empirical formula is how we simplify the whole formula to simplify it to its smallest indivisible parts.
It is definitely not the actual number of atoms. If you see an empirical formula, don't think that it's the full thing.
It is also not a representation of a compound to show its atoms' arrangement: this would be a Lewis dot structure, or a ball and stick model, or something similar. We don't use the empirical formula for this purpose.
A. Fe2O3 + 3CO= 2Fe+3CO2
Here element oxidised is CO or Carbon Monoxide, since oxygen is added.
B. 2HCl+2KMnO4+3H2C2O4=6CO2+2MnO2+2KCl+4H2O
Here Element reduced is 3H2C2O4, since Hydrogen is being added. Also KMnO4 is reduced, since Oxygen is removed.
I answered all of them except 2 for you to do
Hope this helps :))
Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 3.16 L of chlorine were consumed at STP.
Be sure your answer has the correct number of significant digits.
Answer: Thus volume of carbon tetrachloride that would be produced is 0.788 L
Explanation:
According to ideal gas equation:

P = pressure of gas = 1 atm (at STP)
V = Volume of gas = 3.16 L
n = number of moles = ?
R = gas constant =
T =temperature =



According to stoichiometry:
4 moles of chlorine produces = 1 mole of carbon tetrachloride
Thus 0.141 moles of methane produces =
moles of carbon tetrachloride
volume of carbon tetrachloride =
Thus volume of carbon tetrachloride that would be produced is 0.788 L
Answer:
Hydrogen H₂ will be the limiting reagent.
The excess reactant that will be left after the reaction is 3.45 moles.
4.3 moles of water can be produced.
Explanation:
The balanced reation is:
2 H₂ + O₂ → 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- H₂: 2 moles
- O₂: 1 mole
- H₂O: 2 moles
To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂, how much moles of H₂ will be needed if 5.6 moles of O₂ react?

moles of H₂= 11.2 moles
But 11.2 moles of H₂ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of O₂, <u><em>hydrogen H₂ will be the limiting reagent</em></u> and oxygen O₂ will be the excess reagent.
Then you can apply the following rules of three:
- If by reaction stoichiometry 2 moles of H₂ react with 1 mole of O₂, 4.3 moles of H₂ will react with how many moles of O₂?

moles of O₂= 2.15 moles
The excess reactant that will be left after the reaction can be calculated as:
5.6 moles - 2.15 moles= 3.45 moles
<u><em>The excess reactant that will be left after the reaction is 3.45 moles.</em></u>
- If by reaction stoichiometry 2 moles of H₂ produce 2 moles of H₂O, 4.3 moles of H₂ produce how many moles of H₂O?

moles of H₂O= 4.3 moles
<u><em>4.3 moles of water can be produced.</em></u>