Answer:-
29.07 gram of Cu(NO3)2 will be formed.
4.756 grams of AgNO3 will be left over when the reaction is complete.
Explanation:-
Atomic weight of Cu = 63.546 g mol -1
Molecular weight of AgNO3 = 107.87 x 1 + 14 x 1 + 16 x 3
= 169.87 g mol-1
Number of moles of Copper = 9.85 gram / (63.546 g mol-1)
= 0.155 mol
Number of moles of AgNO3 = 31 gram / ( 169.87 g mol-1)
= 0.183 mol
The balanced chemical equation for this reaction is
Cu + AgNO3 --> Cu(NO3)2 + Ag
According to the equation,
1 mole of Cu reacts with 1 mole of AgNO3.
∴0.155 mol of Cu react with 0.155 mol of AgNO3.
Number of moles of AgNO3 left over = 0.183-0.155=0.028 mol
Number of grams of AgNO3 left over = 0.028 mol x 169.87 grams mol-1
= 4.756 gram
Molecular weight of Cu(NO3)2 = 63.546 x 1 + (14 x 1 +16 x 3 ) x 2
=187.546 gram
Now from the balanced chemical equation,
1 Cu gives 1 Cu(NO3)2
∴ 63.546 g of Cu gives 187.546 gram of Cu(NO3)2
9.85 grams pf Cu gives 187.546 x 9.85 / 64.546 gram of Cu(NO3)2
= 29.07 gram of Cu(NO3)2