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scoray [572]
1 year ago
9

What mass of potassium nitrate is needed to generate 215.0 L of gas, composed of 111.0 L of N2 and 104.0 L of O2 at 0.920 atm an

d 291 K, using these two reactions?
Chemistry
1 answer:
pychu [463]1 year ago
6 0

Mass of potassium nitrate is needed to generate 215.0 L of gas, composed of 111.0 L of N₂ and 104.0 L of O₂ at 0.920 atm and 291 K, using these two reactions is 1186.75 g.

The reactions are :

2KNO₃ --->  2KNO₂  +  O₂

4KNO₂  ---->  2K₂O  + 3O₂  +  2N₂

Therefore,

4KNO₃  ---->  2K₂O  +   5O₂  +  2N₂

using the formula:

PV = nRT

moles of N₂ = PV / RT

P = 0.920 atm

T =291 K

R = 0.082 Latm/mol/k

moles of N₂ = 4.279 mol

moles of O₂ = (0.92 × 104) / 0.082 × 291

                   = 4.00 mol

from the equation .

moles of potassium nitrate to produce nitrogen = 2 × 4.279

                                                                                 = 8.55 mol

moles of nitrate tp produce oxygen = (4 / 5) × 4

                                                            = 3.2 mol

total moles of nitrate = 8.55 mol + 3.2 mol

                                   = 11.75 mol

mass of potassium nitrate = number of moles × molar mass

                                           = 11.75 × 101

                                           = 1186.75 g

Thus, Mass of potassium nitrate is needed to generate 215.0 L of gas, composed of 111.0 L of N₂ and 104.0 L of O₂ at 0.920 atm and 291 K, using these two reactions is 1186.75 g.

To learn more about moles here

brainly.com/question/20486415

#SPJ1

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Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

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The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?

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The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield=\frac{actual yield}{theorical yield} x100

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<h3>Percent yield for the reaction in this case</h3>

In this case, you know:

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Learn more about

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brainly.com/question/24653699

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