Chemistry, 14.09.2019 07:30 holycow7868

Determine the percent yield of the following reaction when 2.80 g of p reacts with excess oxygen. the actual yield of this reaction is determined to by 3.89 g of p2o5.
4 p + 5 o2 > 2 p2o5

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Answer from: dannyboy5911

Answer : The percent yield of P_2O_5 is, 30.39 %

Solution : Given,

Mass of P = 2.80 g

Molar mass of P = 31 g/mole

Molar mass of P_2O_5 = 284 g/mole

First we have to calculate the moles of P.

$\text{ Moles of }P=\frac{\text{ Mass of }P}{\text{ Molar mass of }P}=\frac{2.80g}{31g/mole}=0.0903moles$

Now we have to calculate the moles of NH_3

The balanced chemical reaction is,

$4P+5O_2\rightarrow 2P_2O_5$

From the reaction, we conclude that

As, 4 mole of react to give 2 mole of P_2O_5

So, 0.0903 moles of react to give $\frac{0.0903}{4}\times 2=0.04515$ moles of P_2O_5

Now we have to calculate the mass of P_2O_5

$\text{ Mass of }P_2O_5=\text{ Moles of }P_2O_5\times \text{ Molar mass of }P_2O_5$

$\text{ Mass of }P_2O_5=(0.04515moles)\times (284g/mole)=12.8g$

Theoretical yield of P_2O_5 = 12.8 g

Experimental yield of P_2O_5 = 3.89 g

Now we have to calculate the percent yield of P_2O_5

$\% \text{ yield of }P_2O_5=\frac{\text{ Experimental yield of }P_2O_5}{\text{ Theoretical yield of }P_2O_5}\times 100$

$\% \text{ yield of }P_2O_5=\frac{3.89g}{12.8g}\times 100=30.39\%$

Therefore, the percent yield of P_2O_5 is, 30.39 %

Answer from: Quest

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

let g(n) be the amount of contaminants (in grams) that remained by the beginning of the nth hour. g is a sequence, what kind of sequence is it, arithmetic or geometric? then, write an explicit formula for the sequence, g(n)=

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

explanation:

Answer from: Quest

answer; protein;

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