Question

The decomposition of a silver oxide sample forms 14.9 g of Ag(s): 2Ag2O(s)→4Ag(s)+O2(g) What total volume of gas forms if it is collected over water at a temperature of 25 ∘C and a total pressure of 755 mmHg ?

The following reaction consumes 2.30 kg of CO(g): CO(g)+H2O(g)→CO2(g)+H2(g) How many total liters of gas are formed if the products are collected at STP?

Answer 1

Solution-
1)The given equation is
2Ag2O (s) ------------> 4Ag (s) + O2 (g)
From the equation 2 moles of Ag2O decomposes into 4 moles of Ag and 1 mole of O2

4 moles of Ag = 4 * 107.8 = 431.2g of Ag
now at STP, 1 mole of O2 = 22.4 Liters of O2
So at STP
431.2g of Ag decompose into 22.4 Liters of O2
Therefore 14.9 g of Ag will decompose into x Liters of O2

Let’s build a proportion
431.2 / 22.4 = 14.9 / x
x = 0.77 liters of O2 at STP

when collected over water at a temperature of 25 C and a total pressure of 755 mmHg?
water vapor pressure = 23.8 mmHg
gas pressure = 755 – 23.8 = 731.2 mm Hg
temperature = 25°C + 273° = 298° K

(760 * 0.77) / 273 = (731.2 * V lab) / 298

V lab = (760 * 0.77 * 298) ÷ (273 * 731.2)

V lab = 0.87 L

2) The reaction is

CO + H2O → CO2(g) + H2(g)
So;
(2300 g CO) / (28.01021 g CO/mol) x (2 mol gases / 1 mol CO) x (22.414 L/mol)
= 3.68 *10^3 L gases