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.
Learn more: brainly.com/question/22885959
Answer:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Oxidized: Cd
Reduced: Ag
Explanation:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Cd → Cd²⁺ + 2e⁻ Half reaction oxidation
1e⁻ + Ag⁺ → Ag Half reaction reduction
Ag changed oxidation number from +1 to 0
Cd changed oxidation number from 0 to +2
Let's ballance the electrons
( Cd → Cd²⁺ + 2e⁻ ) .1
( 1e⁻ + Ag⁺ → Ag ) .2
Cd + 2e⁻ + 2Ag⁺ → 2Ag + Cd²⁺ + 2e⁻
Finally the ballance equation is:
Cd(s) + 2AgNO₃(aq) → Cd(NO₃)₂ (aq) + 2Ag(s)
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Explanation:
Hope this help...
