2ag2o(s) > 4ag(s) + o2(g)
in the reaction 4.58l of o2 is formed at a pressure of 745 torr and a temperature of 36 c. how many grams of ag2o (s) were required for this much o2 to be produced?

82.03 g de Ag₂O

Explanation:

The ideal gas equation can be used to determine the amount of moles that formed from O₂, and thus be able to calculate the initial amount of Ag₂O.

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 Ag₂O that were needed, and then calculate the mass of them:

1 mole O₂ 2 moles Ag₂O

0.177 mol of O₂ X = 0.354 mol of Ag₂O

Calculation: 0.177 mol x 2 mol / 1 mol = 0.354 mol of Ag₂O

With the help of a periodic table, the molar mass of the compound is calculated

m Ag₂O = 2 x mAg + mO = 2 x 107.87 g + 15.99 g = 231.73 g / mol

1 mole Ag₂O 231.73 g

0.354 mol of Ag₂O X = 82.03 g of Ag₂O

Calculation: 0.354 mol x 231.73 g / 1 mol = 82.03 g of Ag₂O

Therefore, 82.03 g of Ag₂O were needed to form 4.58 L of O₂ at 745 Torr and 36 ° C.

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