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3 years ago

What is the limiting reagent when 49.84 g of nitrogen react with 10.7 g of hydrogen to make ammonia

1 answer:

8 0

If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.

We solve this problem by first writing the equation
N2 + 3H2 = 2NH3
N2 = 14g*2 = 28g, 3H2 = 3(1*2) = 6g
so 28g of Nitrogen needs 6g of Hydrogen for this reaction. Thus if we had 10.67g of Hydrogen in the reaction, 6g*49.84g/28g of hydrogen is needed to react = 10.68g of Hydrogen, but since we have 10.7g of it thus it is excess and thus the limiting reagent has to be Nitrogen, but notice that 10.68g and 10.7g are practically the same, so there might actually not be a limiting reagent. Using the other value(10.7), the amount of Nitrogen required would be 10.7g*28g/6g = 49.93, and since this is slightly more than the 49.84g we have, this confirms that Nitrogen is the limiting reagent. But note still that since this values are really close, there is a possibility that there is neither a limiting nor an excess reagent

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The elements in same period have same principle quantum number or energy shell.

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Diano4ka-milaya [45]

I don't know if this is multiple choice but if it isn't I can name of few.

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3 years ago

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n =

Lady bird [3.3K]

Answer:

$4.86\times10^{-7}\ \text{m}$

Explanation:

R = Rydberg constant = $1.09677583\times 10^7\ \text{m}^{-1}$

$n_1$ = Principal quantum number of an energy level = 2

$n_2$ = Principal quantum number of an energy level for the atomic electron transition = 4

Wavelength is given by the Rydberg formula

$\lambda^{-1}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\\\Rightarrow \lambda^{-1}=1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\\\Rightarrow \lambda=\left(1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\right)^{-1}\\\Rightarrow \lambda=4.86\times10^{-7}\ \text{m}$

The wavelength of the light emitted is $4.86\times10^{-7}\ \text{m}$ .

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3 years ago