1) 3 mole PbS give 2 mole NO
3×239.3 g PbS give 2 mole NO
4.7 g PbS give 2×4.7/3×239.3 g NO
= 0.3928 g NO
= 0.013 mole NO
Using ideal gas equation ,
Volume = nRT/P = 0.013×0.082×300.15/1.1
= 0.29 L
Answer is A)
1. Lead(II) sulfide dissolves in excess nitric acid according to the equation below. Calculate the volume...
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