Net Ionic Practice #2

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Net Ionic Reactions – Practice #2
Grade - -3 each mistake
FYI, here is how each question is scored:
1 point is earned for the correct reactants.
2 points are earned for the correct products.
1 point is earned for correctly balancing the equation for both mass and charge.
1 point is earned for the correct answer or observation for part ii.
Questions:
Questions:
1. (i) A solution of sodium hydroxide is added to a solution of lead(II) nitrate.
(ii) If 1.0 L of 1.0 M solutions of each is mixed together, how many moles of
product will be produced. Assume the reaction goes to completion.
2. (i) Solid magnesium hydroxide is added to a solution of hydrobromic acid.
(ii) What volume, in mL, of 2.00 M hydrobromic acid is required to react
completely with 0.10 mol of solid magnesium hydroxide?
3. (i) Solid strontium hydroxide is added to a solution of nitric acid.
(ii) How many moles of strontium hydroxide would react completely with 500.
mL of 0.40 M nitric acid?
4. (i) Chlorine gas, an oxidizing agent, is bubbled into a solution of potassium
bromide.
(ii) What is the oxidation number of chlorine before the reaction occurs? What
is the oxidation number of chlorine after the reaction occurs?
5. (i) Zinc metal is added to a hydrobromic acid solution.
(ii) Write the oxidation half-reaction for the reaction.
6. (i) A solution of hydrogen peroxide is heated, and a gas is produced.
(ii) Identify the oxidation state of oxygen in hydrogen peroxide.
7. (i) A barium nitrate solution and a potassium fluoride solution are combined and
a precipitate forms.
(ii) If equimolar amounts of barium nitrate and potassium fluoride are
combined, which reactant, if any, is the limiting reactant? Explain.
1. (i) A solution of sodium hydroxide is added to a solution of lead(II) nitrate.
2NaOH + Pb(NO3)2 → 2NaNO3 + Pb(OH)2 (s)
2Na+ + 2OH- + Pb+2 + 2NO3- →2Na+ + 2NO3- + Pb(OH)2
1. 2OH- + Pb+2 → Pb(OH)2
(ii) If 1.0 L of 1.0 M solutions of each is mixed together, how many moles of
product will be produced. Assume the reaction goes to completion.
? π‘šπ‘œπ‘™ 𝑃𝑏(𝑂𝐻)2 = 1.0 𝐿 𝑂𝐻 − π‘₯
1 π‘šπ‘œπ‘™ 1 π‘šπ‘œπ‘™ 𝑃𝑏(𝑂𝐻)2
π‘₯
= 0.5 π‘šπ‘œπ‘™ 𝑃𝑏(𝑂𝐻)2
1𝐿
2 π‘šπ‘œπ‘™ 𝑂𝐻 −
? π‘šπ‘œπ‘™ 𝑃𝑏(𝑂𝐻)2 = 1.0 𝐿 𝑃𝑏 +2 π‘₯
1 π‘šπ‘œπ‘™ 1 π‘šπ‘œπ‘™ 𝑃𝑏(𝑂𝐻)2
π‘₯
= 1.0. π‘šπ‘œπ‘™ 𝑃𝑏(𝑂𝐻)2
1𝐿
1 π‘šπ‘œπ‘™ 𝑃𝑏 +2
2. (i) Solid magnesium hydroxide is added to a solution of hydrobromic acid.
2. Mg(OH)2 + 2 H+ → Mg2+ + 2 H2O
(ii) What volume, in mL, of 2.00 M hydrobromic acid is required to react
completely with 0.10 mol of solid magnesium hydroxide?
? π‘šπΏ π»π΅π‘Ÿ = 0.10 π‘šπ‘œπ‘™ 𝑀𝑔(𝑂𝐻)2 π‘₯
2 π‘šπ‘œπ‘™ π»π΅π‘Ÿ
1 𝐿 π»π΅π‘Ÿ
1000 π‘šπΏ π»π΅π‘Ÿ
π‘₯
π‘₯
= 100 π‘šπΏ
1 π‘šπ‘œπ‘™ 𝑀𝑔(𝑂𝐻)2 2.0 π‘šπ‘œπ‘™ π»π΅π‘Ÿ
𝐿 π»π΅π‘Ÿ
3. (i) Solid strontium hydroxide is added to a solution of nitric acid.
Sr(OH)2 (s) + 2HNO3 οƒ  Sr(NO3)2 + 2HOH
Sr(OH)2 (s) + 2H+ + 2NO3- οƒ  Sr+2 + 2(NO3-) + 2H2O
3. Sr(OH)2 (s) + 2H+ → Sr+2 + 2H2O
(ii) How many moles of strontium hydroxide would react completely with 500.
mL of 0.40 M nitric acid?
? π‘šπ‘œπ‘™ π‘†π‘Ÿ(𝑂𝐻)2 = 500 π‘šπΏ 𝐻𝑁𝑂3 π‘₯
1 𝐿 𝐻𝑁𝑂3
0.40 π‘šπ‘œπ‘™ 𝐻𝑁𝑂3 1 π‘šπ‘œπ‘™ π‘†π‘Ÿ(𝑂𝐻)2
π‘₯
π‘₯
= 0.10 π‘šπ‘œπ‘™
1000 π‘šπΏ 𝐻𝑁𝑂3
1 𝐿 𝐻𝑁𝑂3
2 π‘šπ‘œπ‘™ 𝐻𝑁𝑂3
4. (i) Chlorine gas, an oxidizing agent, is bubbled into a solution of potassium
bromide.
4. Cl2 + 2Br- → 2Cl- + Br2
(ii) What is the oxidation number of chlorine before the reaction occurs? What
is the oxidation number of chlorine after the reaction occurs?
4.ii. Cl2 = 0, Cl- = -1
5. (i) Zinc metal is added to a hydrobromic acid solution.
Zn + 2HBr οƒ  H2 + ZnBr2(aq)
Zn + 2H+ + 2Br- οƒ  H2 + Zn+2 + 2Br5. Zn + 2H+ → H2 + Zn+2
(ii) Write the oxidation half-reaction for the reaction.
5.ii. Zn → Zn+2 + 2 e6. (i) A solution of hydrogen peroxide is heated, and a gas is produced.
6. 2H2O2 → O2 + 2H2O
(ii) Identify the oxidation state of oxygen in hydrogen peroxide.
Normally, oxygen has an oxidation # of -2. This is an exception! Always assign
Hydrogen first, then figure out the rest so that the compound adds up to zero (or
the charge if it is an ion).
O = -1
7. (i) A barium nitrate solution and a potassium fluoride solution are combined and
a precipitate forms.
Ba(NO3)2 + 2 KF οƒ  2 K(NO3) + BaF2
Ba+2 + 2(NO3-) +2 K+ + 2F- οƒ  2 K+ + 2 (NO3-) + BaF2 (s)
7. Ba+2 + 2F- → BaF2
(ii) If equimolar amounts of barium nitrate and potassium fluoride are
combined, which reactant, if any, is the limiting reactant? Explain. KF is the LR.
L.R. depends upon the molar ratio, not on the molarity. You can do a conversion
like this one. I just made up the “1 mol” of reactant part. It could be any number,
so long as they are the same for both reactants (as indicated by the word
equimolar).
? π‘šπ‘œπ‘™ π΅π‘ŽπΉ2 = 1 π‘šπ‘œπ‘™ π΅π‘Ž(𝑁𝑂3 )2 π‘₯
? π‘šπ‘œπ‘™ π΅π‘ŽπΉ2 = 1 π‘šπ‘œπ‘™ 𝐾𝐹 π‘₯
This shows that KF is the LR.
1 π‘šπ‘œπ‘™ π΅π‘ŽπΉ2
= 1 π‘šπ‘œπ‘™
1 π‘šπ‘œπ‘™ π΅π‘Ž(𝑁𝑂3 )2
π‘šπ‘œπ‘™ π΅π‘ŽπΉ2
= 0.5 π‘šπ‘œπ‘™
2 π‘šπ‘œπ‘™ 𝐾𝐹
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