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atomic structure chemistry a level question

a pure sample of 450mg of silver carbonate was decomposed by heating to constant mass. 2ag2co3= 4ag+2co2+o2 calculate the total volume in cm3 of gas produced by 101kpa and 25 degrees. gas constant 8.31

i got the answer as 4x10^-5 but the answer says 6x10^-5cm3 and idk how???
Reply 1
I'm guessing the question wants you to use the ideal gas law.

1.

Find moles of silver carbonate

2.

Find moles of total gas produced (co2 and o2)

3.

Use ideal gas law to find volume of total gas produced

Reply 2
Original post by Methene
I'm guessing the question wants you to use the ideal gas law.

1.

Find moles of silver carbonate

2.

Find moles of total gas produced (co2 and o2)

3.

Use ideal gas law to find volume of total gas produced


I did do that but I got a different answer
Reply 3

1.

n silver carbonate = 0.45/(275.8) = 1.63 x 10-3

2.

n gas produced = (1.63 x 10-3) x 3/2 = 2.45 x 10-3

3.

Vol. gas produced = (2.45 x 10-3) x 8.31 x 298 / 101000 = 6 x 10-5

Original post by Riema
a pure sample of 450mg of silver carbonate was decomposed by heating to constant mass. 2ag2co3= 4ag+2co2+o2 calculate the total volume in cm3 of gas produced by 101kpa and 25 degrees. gas constant 8.31
i got the answer as 4x10^-5 but the answer says 6x10^-5cm3 and idk how???

Relative mass of silver carbonate = 2(107.87) + 60 = 168
450mg = 0.45g = 0.45/275.74 mol = 1.63 x 10-3
2Ag2CO3 ==> 4Ag + 2CO2 + O2
Hence 2 mol of silver carbonate makes 3 mol gas
Hence 1 mol of silver carbonate makes 1.5 mol gas
Hence mol of gas = 1.5 x 1.63 x 10-3 = 2.45 x 10-3
PV = nRT
V = nRT/P = 2.45 x 10-3 x 8.314 x 298/101 = 0.06 dm3 = 60cm3
Original post by Methene

1.

n silver carbonate = 0.45/(275.8) = 1.63 x 10-3

2.

n gas produced = (1.63 x 10-3) x 3/2 = 2.45 x 10-3

3.

Vol. gas produced = (2.45 x 10-3) x 8.31 x 298 / 101000 = 6 x 10-5


Your units are not coherent. If you use P in pascals then your answer for V is in m3
Reply 6
Original post by charco
Your units are not coherent. If you use P in pascals then your answer for V is in m3

Yep I meant m3 in my answer. Didn't notice that the answer OP stated is in cm3, so I suppose that's either a typo on their part or the MS?
You just calculated the volume of CO2 produced, question is asking for volume of gas produced, both CO2 and O2 gas are produced in here
Reply 8
Original post by Methene

1.

n silver carbonate = 0.45/(275.8) = 1.63 x 10-3

2.

n gas produced = (1.63 x 10-3) x 3/2 = 2.45 x 10-3

3.

Vol. gas produced = (2.45 x 10-3) x 8.31 x 298 / 101000 = 6 x 10-5


thank you i get it now!!!
Reply 9
Original post by charco
Relative mass of silver carbonate = 2(107.87) + 60 = 168
450mg = 0.45g = 0.45/275.74 mol = 1.63 x 10-3
2Ag2CO3 ==> 4Ag + 2CO2 + O2
Hence 2 mol of silver carbonate makes 3 mol gas
Hence 1 mol of silver carbonate makes 1.5 mol gas
Hence mol of gas = 1.5 x 1.63 x 10-3 = 2.45 x 10-3
PV = nRT
V = nRT/P = 2.45 x 10-3 x 8.314 x 298/101 = 0.06 dm3 = 60cm3

thank you!!!

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