Consider the unbalanced equation for the reaction of solid lead with silver nitrate:
\(\mathrm{Pb}(\mathrm{s})+\mathrm{AgNO}_{3}(a q) \rightarrow \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Ag}(s)\)
(a) Balance the equation.
(b) How many moles of silver nitrate are required to completely react with 9.3 mol of lead?
(c) How many moles of Ag are formed by the complete reaction of 28.4 mol of Pb?
Equation Transcription:
Text Transcription:
Pb(s) + AgNO_3 (aq) right arrow Pb(NO_3)_2 (aq) + Ag(s)
Solution 29P
Step 1:
(a)The above equation can be balanced as follows,
Pb(s) + AgNO3(aq) → Pb(NO3)2(aq) + Ag(s)
In the above chemical equation, the number of mole of NO3 is unequal, in order to balance the total number of NO3, multiply 2 with AgNO3 and number of silver atoms can also be balanced by multiplying 2 with Ag(s).
Thus the balanced equation will be,
Pb(s) + 2AgNO3(aq) → Pb(NO3)2(aq) + 2Ag(s)
