Question

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 and 291 K, using these two reactions?

Answer 1

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.

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