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3 years ago

Write a equilibrium constant expression for the following equilibrium equation

Chemistry

1 answer:

elena55 [62]3 years ago

7 0

Answer:[co]^2/[co2]

Explanation:

kc=[products^no of moles]/[reactants^no of moles]

Note: solid and liquid water cant be taken in the kc calculations because it has constant concentration

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2 years ago

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2 years ago

If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.

kiruha [24]

Answer:

Sr(OH)₂ will be the limiting reagent.

Explanation:

First of all, you should know the following balanced chemical equation:

2 H₂CO₃ + 2 Sr(OH)₂ → 4 H₂O + Sr₂(CO₃)₂

The balanced equation is based on the Law of Conservation of Mass, which says that matter cannot be created or destroyed. Therefore, the number of each type of atom on each side of a chemical equation must be the same.

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). By stoichiometry the following amounts in moles react:

strontium hydroxide: 2 moles
carbonic acid: 2 moles

Now, you know the following masses of the elements:

Sr: 87.62 g/mole
O: 16 g/mole
H: 1 g/mole

So the molar mass of strontium hydroxide is:

Sr(OH)₂= 87.62 g/mole + 2*(16 g/mole + 1 g/mole)= 121.62 g/mole

You apply the following rule of three, if 121.62 grams of hydroxide are present in 1 mole, 12.5 grams in how many moles are they?

$moles of strontium hydroxide=\frac{12.5 grams*1 mole}{121.62 grams}$

moles of hydroxide= 0.103 moles

On the other hand, you have 150 ml of 3.5 M carbonic acid. Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you can apply the following rule of three: if in 1 L there are 3.5 moles of carbonic acid, in 0.150 L (being 1 L = 1000 mL, 0.150 L = 150 mL) how many moles of acid are there?

$molesofcarbonicacid=\frac{0.150 L*3.5 moles}{1 L}$

moles of carbonic acid= 0.525 moles

Finally, to calculate the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 2 mole of strontium hydroxide reacts with , how much moles of carbonic acid will be needed if 0.103 moles of strontium hydroxide react?

$molesofcarbonicacid=\frac{0.103 moles of strontium hydroxide*2 moles of carbonic acid}{2 moles of strontium hydroxide}$

moles of carbonic acid= 0.103 moles

But 0.525 moles are available. Since more moles are available than you need to react with 0.103 moles of strontium hydroxide, <u><em>Sr(OH)₂ will be the limiting reagent.</em></u>

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3 years ago