CHEMISTRY | MOLE CONCEPT | Question 25
The decomposition of Pb(NO3)2 is as shown below.
Pb(NO3)2(S)→2PbO(S) + 4NO2(g) + O2(g)
If 0.01 mole of the nitrate was used, what would be the maximum volume of oxygen gas collected at room temperature and pressure?
(molar gas volume at rtp = 24.0 cm33)
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Data
Mole of Pb(NO3)2 = 0.01 mole
Asked:
Volume of oxygen gas collected.
To use mole equivalent
1 mole of Pb(NO3)2 Ξ 1 mole O2
0.01 mole Pb(NO3)2 Ξ ?? O2
= (0.01 mole × 1mole) / 1 mole
= 0.1 mole O2
Volume of oxygen will be:
Volume = mole × molar volume
= 0.01 mole × 22.4 dm3/mole
= 0.224dm3
0.224dm3 of gas was collected at SLP
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