In the following reaction, 4.58 L of O2 was formed at P = 745 mmHg and T = 308 K.?

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2017-10-20 2:41 pm

In the following reaction, 4.58 L of O2 was formed at P = 745 mmHg and T = 308 K. How many grams of Ag2O decomposed? (Ag2O = 231.735 g/mol)2Ag2O(s) →4 Ag (s) + O2(g)

I know the answer is 82.29 G.... please explain how you can get to that answer!!

回答 (1)

wanszeto

2017-10-20 3:56 pm

✔ 最佳答案

Consider the O₂ formed in the reaction.
Pressure, P = 745 mmHg = 745/760 atm
Volume, V = 4.58 L
Gas constant, R = 0.0821 atm L / (mol K)
Temperature, T = 308 K

Gas law: PV = nRT
Then, n = PV/(RT)
No. of moles of O₂ formed, n = (745/760) × 4.58 / (0.0821 × 308) = 0.17755 mol

Consider the following reaction.
2Ag₂O(s) → 4Ag(s) + O₂(g)
Mole ratio Ag₂O : O₂ = 2 : 1
No. of moles of Ag₂O decomposed = (0.17755 mol) × 2 = 0.3551 mol

Molar mass of Ag₂O = 231.735 g/mol
Mass of Ag₂O decomposed = (0.3551 mol) × (231.735 g/mol) =82.29 g

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