One part nitrogen gas combines with one part oxygen gas to form how many part(s) dinitrogen monoxide (nitric oxide)?
1 answer:
Answer : The one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.
Explanation :
Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.
When nitrogen gas combines with oxygen gas then it react to give dinitrogen monoxide or nitrous oxide.
The balance chemical reaction will be:
By the stoichiometry we can say that, 2 parts of nitrogen gas combines with 1 part of oxygen gas to give 2 parts of dinitrogen monoxide or nitrous oxide.
First we have to determine the limiting reagent.
From the reaction we conclude that,
As, 2 moles of nitrogen gas combine with 1 mole of oxygen gas
So, 1 moles of nitrogen gas combine with 0.5 mole of oxygen gas
It means that, oxygen gas is an excess reagent because the given moles are greater than the required moles and nitrogen gas is a limiting reagent and it limits the formation of product.
Now we have to determine the moles of dinitrogen monoxide.
As, 2 moles of nitrogen gas combine to give 2 mole of dinitrogen monoxide
So, 1 mole of nitrogen gas combine to give 1 mole of dinitrogen monoxide
Thus, the one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.
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Without wasting much of our time, Here is the correct question.
Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemoglobin(aq) + O2(aq) -------> hemoglobin O2(aq) is first order in hemoglobin and first order in dissolved oxygen, with a rate constant of 4 × 10⁷ L mol⁻¹ s⁻¹. Calculate the initial rate at which oxygen will be bound to hemoglobin if the concentration of hemoglobin is 2 × 10⁻⁹ M and that of oxygen is 5 × 10⁻⁵M.
Answer:
4 × 10⁻⁶ M s⁻¹
Explanation:
The equation for the reaction between Hemoglobin molecules in blood that binds with oxygen molecule can be represent by:
+
--------->
Now, we are also being told to calculate only!, the initial rate at which oxygen will be bound to hemoglobin.
So, If it is first order in hemoglobin and also first order in Oxygen molecule at the initial rate of the the reaction, therefore, the rate for the reaction can be expressed as :
rate = k
Let's not forget that we are so given some parameters;
where
k (rate constant) = 4 × 10⁷ L mol⁻¹ s⁻¹
[ ] = 2 × 10⁻⁹ M
[ ] = 5 × 10⁻⁵ M
Substituting our data given into the above rate formula, we have:
rate = (4 × 10⁷ L mol⁻¹ s⁻¹) × (2 × 10⁻⁹ M) × (5 × 10⁻⁵ M)
rate = 4 × 10⁻⁶ M s⁻¹ ( given that 1 M = 1 mol L⁻¹ )
∴ the initial rate at which oxygen will be bound to hemoglobin = 4 × 10⁻⁶ M s⁻¹