Asked by Shivtej

consider the following reaction AgNO3+KBr=AgBr+KNO3
calculate the amount of presipitate in gram obtained when 0.25 mole of AgNO3 is trheated with excess of aqueous KBr solution
Answered by Steve
you get 0.25 moles of AgBr, so what is its mass?
Answered by sd
since their is 0.25 moles of AgNO3 hence you the mass = 42.5g and similarly mass of kbr =42.5 g or 0.35 moles by GM/MM = NO OF MOLES
hence by this we can get that their is 0.25 moles of AgBr thier for its mass is 47 g by applying law of conservation of mass
Answered by Madhusudhan Koda
AgNO3 + KBr → AgBr ↓ + KNO3
1M 1M
0.25M 0.25M
0.25M of AgBr=108+80=188/4=47
Ans: 47
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