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
3 answers
you get 0.25 moles of AgBr, so what is its mass?
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
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
AgNO3 + KBr → AgBr ↓ + KNO3
1M 1M
0.25M 0.25M
0.25M of AgBr=108+80=188/4=47
Ans: 47
1M 1M
0.25M 0.25M
0.25M of AgBr=108+80=188/4=47
Ans: 47
1 answer