
How many grams of $ {{{O}}_{{2}}} $ are needed to react with $ {{84}}{{.5}} $ g of $ {{N}}{{{H}}_{{3}}} $ ?
Answer
549.9k+ views
Hint: In the above question, we are asked to find out weight of oxygen which are needed to react with $ {{84}}{{.5}} $ g of $ {{N}}{{{H}}_{{3}}} $ . For this first, we have to write a balanced chemical equation of reaction of ammonia with oxygen. Then we can easily infer how much moles of $ {{N}}{{{H}}_{{3}}} $ reacts with how much oxygen. The next step will be finding out the number of moles of $ {{N}}{{{H}}_{{3}}} $ , consequently, the number of moles and weight of oxygen can be found out.
Formula Used
$ {{n = }}\dfrac{{{m}}}{{{M}}} $
Where n is the number of moles
m = given mass
M = molar mass.
Complete step by step solution
Let us write a balanced chemical reaction of $ {{N}}{{{H}}_{{3}}} $ and $ {{{O}}_{{2}}} $ .
$ {{4N}}{{{H}}_{{3}}}{{ + 5}}{{{O}}_{{2}}} \to {{4NO + 6}}{{{H}}_{{2}}}{{O}} $ ..........(1)
We can see that 4 moles of ammonia react with 5 moles of oxygen.
Let us find the number of moles of $ {{N}}{{{H}}_{{3}}} $ present.
Molar mass of $ {{N}}{{{H}}_{{3}}} $ = atomic mass of N + $ {{3 \times }} $ atomic mass of H = $ {{14 + 3 \times 1 = 17}} $ g
Hence, $ {{n = }}\dfrac{{{m}}}{{{M}}}{{ = }}\dfrac{{{{84}}{{.5}}}}{{{{17}}}}{{ = 4}}{{.97}} $
Since, 4 moles of $ {{N}}{{{H}}_{{3}}} $ reacts with 5 moles of $ {{{O}}_{{2}}} $ . (From 1)
Hence, $ {{4}}{{.97}} $ moles of $ {{N}}{{{H}}_{{3}}} $ reacts with $ \dfrac{{{5}}}{{{4}}}{{ \times 4}}{{.97 = 6}}{{.21}} $ moles of $ {{{O}}_{{2}}} $ .
Let us now find out the molar mass of $ {{{O}}_{{2}}} $ .
Molar mass of $ {{{O}}_{{2}}} $ = $ {{2 \times }} $ atomic mass of O = $ {{2 \times 16 = 32}} $ g of $ {{{O}}_{{2}}} $
We know that:
$ {{n = }}\dfrac{{{m}}}{{{M}}} $
Rearranging the above equation, we get:
$ {{m = n \times M}} $
Substituting the values of number of moles and molar mass of oxygen, we get:
$ {{m = n \times M = 6}}{{.21 \times 32 = 198}}{{.8}} $ g
So, $ {{198}}{{.8}} $ g of oxygen reacts with $ {{84}}{{.5}} $ g of $ {{N}}{{{H}}_{{3}}} $ .
Note
An equation must be balanced in order to satisfy the law of conservation of mass which states that mass can neither be created nor be destroyed.
If an equation is not balanced, then nothing can be inferred from the equation.
Formula Used
$ {{n = }}\dfrac{{{m}}}{{{M}}} $
Where n is the number of moles
m = given mass
M = molar mass.
Complete step by step solution
Let us write a balanced chemical reaction of $ {{N}}{{{H}}_{{3}}} $ and $ {{{O}}_{{2}}} $ .
$ {{4N}}{{{H}}_{{3}}}{{ + 5}}{{{O}}_{{2}}} \to {{4NO + 6}}{{{H}}_{{2}}}{{O}} $ ..........(1)
We can see that 4 moles of ammonia react with 5 moles of oxygen.
Let us find the number of moles of $ {{N}}{{{H}}_{{3}}} $ present.
Molar mass of $ {{N}}{{{H}}_{{3}}} $ = atomic mass of N + $ {{3 \times }} $ atomic mass of H = $ {{14 + 3 \times 1 = 17}} $ g
Hence, $ {{n = }}\dfrac{{{m}}}{{{M}}}{{ = }}\dfrac{{{{84}}{{.5}}}}{{{{17}}}}{{ = 4}}{{.97}} $
Since, 4 moles of $ {{N}}{{{H}}_{{3}}} $ reacts with 5 moles of $ {{{O}}_{{2}}} $ . (From 1)
Hence, $ {{4}}{{.97}} $ moles of $ {{N}}{{{H}}_{{3}}} $ reacts with $ \dfrac{{{5}}}{{{4}}}{{ \times 4}}{{.97 = 6}}{{.21}} $ moles of $ {{{O}}_{{2}}} $ .
Let us now find out the molar mass of $ {{{O}}_{{2}}} $ .
Molar mass of $ {{{O}}_{{2}}} $ = $ {{2 \times }} $ atomic mass of O = $ {{2 \times 16 = 32}} $ g of $ {{{O}}_{{2}}} $
We know that:
$ {{n = }}\dfrac{{{m}}}{{{M}}} $
Rearranging the above equation, we get:
$ {{m = n \times M}} $
Substituting the values of number of moles and molar mass of oxygen, we get:
$ {{m = n \times M = 6}}{{.21 \times 32 = 198}}{{.8}} $ g
So, $ {{198}}{{.8}} $ g of oxygen reacts with $ {{84}}{{.5}} $ g of $ {{N}}{{{H}}_{{3}}} $ .
Note
An equation must be balanced in order to satisfy the law of conservation of mass which states that mass can neither be created nor be destroyed.
If an equation is not balanced, then nothing can be inferred from the equation.
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