Answer
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Hint: Before solving the question, you must know what limiting reagent is, the limiting reagent is the chemical substance in the reaction which is totally consumed when the reaction is completed.
Complete step by step solution:
First thing you need to know is the molar mass of 1 mole of Oxygen and Carbon. So that, you can calculate the molar mass of the compound.
C + \[{{O}_{2}}\]= \[C{{O}_{2}}\]
Molar Mass of Oxygen = 16g
So, the dioxygen will be 16 X 2 = 32g
Molar mass of carbon = 12g
Molar mass of \[C{{O}_{2}}\]= 12 + 32 = 44g
We can conclude from the equation when 1 mole of carbon burnt in 1 mole of dioxygen (molecular mass = 32), gives rise to 1 mole of carbon dioxide (molecular mass=44).
Suppose, when 32g of oxygen (equal to 1 mole of \[{{O}_{2}}\]) is heated gives \[C{{O}_{2}}\]= 44g
So 1g of oxygen upon heating will give = \[\dfrac{44}{32}\]
Therefore, 16g of dioxygen upon heating will give CO2 = \[\dfrac{44}{32}\] X 16= 22g
Now according to the question only 16g of dioxygen is available for the reaction.
Number of moles of Oxygen = \[\dfrac{16}{32}\]= 0.5 moles.
Thus, Oxygen will act as the limiting reagent.
So, even when 2 moles of carbon are available for reaction, only 0.5 moles of oxygen will be available to combine with carbon.
Amount of \[C{{O}_{2}}\] produced will be 0.5 X 44 = 22g which is the same as the previous example.
So, from the above statement we can conclude that the mass of \[C{{O}_{2}}\] produced is 22g.
Note: Always calculate the number of moles of reactants in order to figure out the limiting reagent.
We can do this by:
Firstly, balance the chemical equation.
Then we will convert them into moles (most likely, through the use of molar mass as a conversion factor).
First calculate the mole ratio from the given information and then compare the calculated ratio to the actual ratio.
Now we can use the amount of limiting reactant to calculate the amount of product produced.
If asked, also calculate how much of the non-limiting reagent is left.
Complete step by step solution:
First thing you need to know is the molar mass of 1 mole of Oxygen and Carbon. So that, you can calculate the molar mass of the compound.
C + \[{{O}_{2}}\]= \[C{{O}_{2}}\]
Molar Mass of Oxygen = 16g
So, the dioxygen will be 16 X 2 = 32g
Molar mass of carbon = 12g
Molar mass of \[C{{O}_{2}}\]= 12 + 32 = 44g
We can conclude from the equation when 1 mole of carbon burnt in 1 mole of dioxygen (molecular mass = 32), gives rise to 1 mole of carbon dioxide (molecular mass=44).
Suppose, when 32g of oxygen (equal to 1 mole of \[{{O}_{2}}\]) is heated gives \[C{{O}_{2}}\]= 44g
So 1g of oxygen upon heating will give = \[\dfrac{44}{32}\]
Therefore, 16g of dioxygen upon heating will give CO2 = \[\dfrac{44}{32}\] X 16= 22g
Now according to the question only 16g of dioxygen is available for the reaction.
Number of moles of Oxygen = \[\dfrac{16}{32}\]= 0.5 moles.
Thus, Oxygen will act as the limiting reagent.
So, even when 2 moles of carbon are available for reaction, only 0.5 moles of oxygen will be available to combine with carbon.
Amount of \[C{{O}_{2}}\] produced will be 0.5 X 44 = 22g which is the same as the previous example.
So, from the above statement we can conclude that the mass of \[C{{O}_{2}}\] produced is 22g.
Note: Always calculate the number of moles of reactants in order to figure out the limiting reagent.
We can do this by:
Firstly, balance the chemical equation.
Then we will convert them into moles (most likely, through the use of molar mass as a conversion factor).
First calculate the mole ratio from the given information and then compare the calculated ratio to the actual ratio.
Now we can use the amount of limiting reactant to calculate the amount of product produced.
If asked, also calculate how much of the non-limiting reagent is left.
Watch videos on
Calculate the amount of \[C{{O}_{2}}\] that could be produced when 2 moles of carbon are burnt in 16g of dioxygen.
Some Basic Concepts of Chemistry | NCERT EXERCISE 1.4 | Class 11 Chemistry Chapter 1 | Nandini Ma'am
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