Answer:
stochiometry works with measuring quantitative relationships and used to determine the amount of products and reactants that are produced or needed in a reaction
Answer:
I think the answers D. Hope this helps you.
Explanation:
Answer:
The Kc of this reaction is 311.97
Explanation:
Step 1: Data given
Kp = 0.174
Temperature = 243 °C
Step 2: The balanced equation
N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 3: Calculate Kc
Kp = Kc *(RT)^Δn
⇒ with Kp = 0.174
⇒ with Kc = TO BE DETERMINED
⇒ with R = the gas constant = 0.08206 Latm/Kmol
⇒ with T = the temperature = 243 °C = 516 K
⇒ with Δn = number of moles products - moles reactants 2 – (1 + 3) = -2
0.174 = Kc (0.08206*516)^-2
Kc = 311.97
The Kc of this reaction is 311.97
Sodium(Na) is the limiting reagent.
<h3>What is Limiting reagent?</h3>
The reactant that is totally consumed during a reaction, or the limiting reagent, decides when the process comes to an end. The precise quantity of reactant required to react with another element may be estimated from the reaction stoichiometry.
How do you identify a limiting reagent?
The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced. Calculate how many moles of each reactant are present and contrast this ratio with the mole ratio of the reactants in the balanced chemical equation to get the limiting reactant.
Start by writing the balanced chemical equation that describes this reaction

Notice that the reaction consumes 2 moles of sodium metal for every 1 mole of chlorine gas that takes part in the reaction and produces 2 moles of sodium chloride.
now we can see that we have 3 moles of sodium and 3 moles of chlorine, according to question. so, we can say that sodium is the limiting reagent in the given situation.
to learn more about Limiting Reagent go to - brainly.com/question/14222359
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In a metallic bond, atoms of the metal are surrounded by a constantly moving "sea of electrons". This moving sea of electrons enables the metal to conduct electricity and move freely among the ions.