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Juli2301 [7.4K]
3 years ago
10

Consider the equations below. (1) Ca(s) + CO2(g) + 1 2 O2(g) → CaCO3(s) (2) 2Ca(s)+O2(g) → 2CaO(s) How should you manipulate the

se equations so that they produce the equation below when added? Check all that apply. CaO(s) + CO2(g) → CaCO3(s) reverse the direction of equation (2) multiply equation (1) by 3 multiply equation (2) by 1/2
Chemistry
1 answer:
In-s [12.5K]3 years ago
3 0

Answer : The correct options are, reverse the direction of equation (2) and multiply equation (2) by 1/2.

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The main given balanced chemical reaction is,

CaO(s)+CO_2(g)\rightarrow CaCO_3(s)

Now we have to determine the main chemical reaction from the given two intermediate reactions.

(1) Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)

(2) 2Ca(s)+O_2(g)\rightarrow 2CaO(s)

Now reverse the equation (2), multiply equation (2) by 1/2 and then add both the equations, we get the main chemical reaction.

(1) Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)

(2) CaO(s)\rightarrow Ca(s)+\frac{1}{2}O_2(g)

Now add both the equations, we get:

CaO(s)+CO_2(g)\rightarrow CaCO_3(s)

Hence, the steps used for the main reaction are, reverse the direction of equation (2) and multiply equation (2) by 1/2.

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How many moles of NH3 would be formed from the complete reaction of 16.0 g H2?
natima [27]

Taking into account the reaction stoichiometry, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

N₂ + 3 H₂ → 2 NH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • N₂: 1 mole
  • H₂: 3 moles
  • NH₃: 2 moles

The molar mass of the compounds is:

  • N₂: 14 g/mole
  • H₂: 2 g/mole
  • NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • N₂: 1 mole ×14 g/mole= 14 grams
  • H₂: 3 moles ×2 g/mole= 6 grams
  • NH₃: 2 moles ×17 g/mole=34 grams

<h3>Mass of NH₃ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 2 moles of NH₃, 16 grams of H₂ form how many moles of NH₃?

moles of NH_{3}= \frac{16 grams of H_{2} x2moles of NH_{3}}{6 grams of H_{2}}

<u><em>moles of NH₃= 5.33 moles</em></u>

Then, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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