There are 4 moles of spectator ions that remain in solution.
The equation of the reaction is;
Na2CO3(aq) + Pb(NO3)2(aq) -------> PbCO3(s) + 2NaNO3(aq)
We have to determine the limiting reactant. This is the reactant that yields the least amount of product. Note that the spectator ions are Na^+ and NO3^- that form NaNO3.
For Na2CO3
1 mole of Na2CO3 yields 2 moles of NaNO3
3 moles of Na2CO3 yields 3 × 2/1 = 6 moles of NaNO3
For Pb(NO3)2
1 mole of Pb(NO3)2 yields 2 moles of NaNO3
2 moles of Pb(NO3)2 yields 2 × 2/1 = 4 moles of NaNO3
We can see that Pb(NO3)2 is the limiting reactant.
Since [NaNO3] = [Na^+] = [NO3^-], it follows that there are 4 moles of spectator ions that remain in solution.
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Answer:
Tile 52 and three are incorrect
Explanation:
Right now it's written in scientific notation, so you can just move the decimal place in 3.8 to the right 3 times (as it is times 10 to the third power) to get 3,800g.
Answer:
in both nucleophil attach the c and leaving group leave but in acyl nu. subsituation c of carbonyl because of double bond with o have bigger positive charge and is better electrophil so do it faster,also alkyl nu. subsituation can have rearangment if going from sn1 and in sn2 sterichemistry of molecule change , acyl nu. subsituation most of time is better