Answer:
82.03 g de Agâ‚‚O
Explanation:
The ideal gas equation can be used to determine the amount of moles that formed from Oâ‚‚, and thus be able to calculate the initial amount of Agâ‚‚O.
Knowing that the equation is:
PxV = n x R x T ⇒ n = (PxV) / (RxT)
Where:
• P: Gas pressure in atm
• V: Volume of gas in L
• n: Number of moles of gas
• R: Constant of ideal gases (0.08206 L.atm / mol.K)
• T: Temperature in K
The corresponding unit conversions must be performed and then replaced in the equation to solve it, then:
Pressure:
760 Torr _____ 1 atmosphere
745 Torr _____ X = 0.98 atm
Calculation: 745 Torr x 1 atm / 760 Torr = 0.98 atm
Temperature:
T (K) = t (° C) + 273.15 = 36 ° C +273.15 = 309.15 K
n = (PxV) / (RxT) = (0.98 atm x 4.58L) / (0.08206 L.atm / mol.K x 309.15 K) = 0.176925 ≅ 0.177 mol of O₂
Knowing the amount of oxygen that formed in moles, we can calculate the initial moles of Agâ‚‚O that were needed, and then calculate the mass of them:
1 mole Oâ‚‚ _____ 2 moles Agâ‚‚O
0.177 mol of Oâ‚‚ _____ X = 0.354 mol of Agâ‚‚O
Calculation: 0.177 mol x 2 mol / 1 mol = 0.354 mol of Agâ‚‚O
With the help of a periodic table, the molar mass of the compound is calculated
m Agâ‚‚O = 2 x mAg + mO = 2 x 107.87 g + 15.99 g = 231.73 g / mol
1 mole Agâ‚‚O _____ 231.73 g
0.354 mol of Agâ‚‚O _____ X = 82.03 g of Agâ‚‚O
Calculation: 0.354 mol x 231.73 g / 1 mol = 82.03 g of Agâ‚‚O
Therefore, 82.03 g of Ag₂O were needed to form 4.58 L of O₂ at 745 Torr and 36 ° C.
