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.
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Answer:
12.32 L.
Explanation:
The following data were obtained from the question:
Mass of CH4 = 8.80 g
Volume of CH4 =?
Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:
Mass of CH4 = 8.80 g
Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol
Mole of CH4 =?
Mole = mass/Molar mass
Mole of CH4 = 8.80 / 16
Mole of CH4 = 0.55 mole.
Finally, we shall determine the volume of the gas at stp as illustrated below:
1 mole of a gas occupies 22.4 L at stp.
Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.
Thus, 8.80 g of CH4 occupies 12.32 L at STP.
Answer:
Hello
12 grams
The mass of one mole of carbon-12 atoms is 12 grams.
Hope it helps You.....
Explanation:
Answer:
kp= 3.1 x 10^(-2)
Explanation:
To solve this problem we have to write down the reaction and use the ICE table for pressures:
2SO2 + O2 ⇄ 2SO3
Initial 3.4 atm 1.3 atm 0 atm
Change -2x - x + 2x
Equilibrium 3.4 - 2x 1.3 -x 0.52 atm
In order to know the x value:
2x = 0.52
x=(0.52)/2= 0.26
2SO2 + O2 ⇄ 2SO3
Equilibrium 3.4 - 0.52 1.3 - 0.26 0.52 atm
Equilibrium 2.88 atm 1.04 atm 0.52 atm
with the partial pressure in the equilibrium, we can obtain Kp.
