Plz Help ASAP please

Question

Silver oxide is heated, and the oxygen is driven off and solid silver is formed. Determine the mass percent of oxygen in Ag2O based on the following experimental data:

2 Ag2O (s) → 4 Ag (s) + O2 (g)

Mass of test tube 40.796 g

Mass of test tube + Ag2O 49.123g

Mass of test tube + Ag 48.561g

A. Calculate the starting mass of Ag2O

B. Calculate the mass of oxygen released

C. Calculate the experimental percentage of oxygen released

D. Calculate the accepted value of percent oxygen in Ag2O (for this determine % oxygen in the silver oxide formula)

E. Calculate the percent error

Answer 1

Part a)

Mass of Ag₂O = (Mass of test tube + Ag₂O) - (Mass of test tube)

Mass of Ag₂O = (49.123 g) - (40.796 g)

The mass of Ag₂O is 8.327. g.

Part b)

Mass of O_{​​​​​​2} = (Mass of test tube + Ag₂O) - (Mass of test tube + Ag)

Mass of O_{​​​​​​2} = (49.123 g) - (48.561 g)

The mass of Ag₂O is 0.562 g.

Part c)

The experimental percentage of oxygen = (Mass of oxygen / Mass of Ag₂O) *100

The experimental percentage of oxygen = (0.562/8.327)*100.

The experimental percentage of oxygen = 6.75%

Part d)

Molar mass of Ag₂O = 231.735g

In one mole of Ag₂O, there is 1 mole of oxygen, i.e., 16g

The accepted theoretical value of percent oxygen

= (16 /231.735)*100 = 6.90%

Part e)

Now percentage error =( accepted value-theoretical value)/theoretical value * 100

= {(6.90 - 6.75)/6.90 } *100

= 2.174%