Question

directions: for the problems below identify the missing reactants or products and then balance the resulting equation.

1. mg + ____ → mgo
2. li + cl₂ → ____
3. ____ + o₂ → fe₂o₃
4. ca + ____ → cal

extra credit challenge: balance each of the following equations. you must show your work for credit.

1. __pb(no₃)₂(aq) + naoh(aq) → pb(oh)₂(s) + __nano₃(aq)
2. __c₈h₁₈(g) + o₂(g) → co₂(g) + __h₂o(g)

Explanation:

Step 1: Identify the reaction type (Problem 4)

The reaction is between magnesium (Mg) and oxygen (O₂) to form magnesium oxide (MgO). First, write the unbalanced equation: $ \text{Mg} + \text{O}_2
ightarrow \text{MgO} $

Step 2: Balance the oxygen atoms

On the left, there are 2 O atoms (from O₂), and on the right, 1 O atom (from MgO). To balance O, put a coefficient of 2 in front of MgO: $ \text{Mg} + \text{O}_2
ightarrow 2\text{MgO} $

Step 3: Balance the magnesium atoms

Now, there are 2 Mg atoms on the right (from 2 MgO), so put a coefficient of 2 in front of Mg on the left: $ 2\text{Mg} + \text{O}_2
ightarrow 2\text{MgO} $

Step 1: Identify the reaction type (Problem 5)

The reaction is between iron (Fe) and oxygen (O₂) to form iron(III) oxide (Fe₂O₃). Unbalanced equation: $ \text{Fe} + \text{O}_2
ightarrow \text{Fe}_2\text{O}_3 $

Step 2: Balance the oxygen atoms

Find the least common multiple of 2 (O₂) and 3 (Fe₂O₃), which is 6. So, put 3 in front of O₂ (3×2=6 O) and 2 in front of Fe₂O₃ (2×3=6 O): $ \text{Fe} + 3\text{O}_2
ightarrow 2\text{Fe}_2\text{O}_3 $

Step 3: Balance the iron atoms

Now, there are 4 Fe atoms on the right (2×2), so put 4 in front of Fe on the left: $ 4\text{Fe} + 3\text{O}_2
ightarrow 2\text{Fe}_2\text{O}_3 $

Step 1: Identify the reaction type (Problem 6)

Lithium (Li) reacts with chlorine (Cl₂) to form lithium chloride (LiCl). Unbalanced equation: $ \text{Li} + \text{Cl}_2
ightarrow \text{LiCl} $

Step 2: Balance the chlorine atoms

There are 2 Cl atoms on the left (Cl₂), so put 2 in front of LiCl: $ \text{Li} + \text{Cl}_2
ightarrow 2\text{LiCl} $

Step 3: Balance the lithium atoms

Now, 2 Li on the right, so put 2 in front of Li on the left: $ 2\text{Li} + \text{Cl}_2
ightarrow 2\text{LiCl} $

Step 1: Identify the reaction type (Problem 7)

Calcium (Ca) reacts with iodine (I₂) to form calcium iodide (CaI₂). Unbalanced equation: $ \text{Ca} + \text{I}_2
ightarrow \text{CaI}_2 $

Step 2: Check balance

Ca: 1 on left, 1 on right. I: 2 on left (I₂), 2 on right (CaI₂). So it's already balanced? Wait, no—wait, the product is CaI₂. Wait, the unbalanced equation is $ \text{Ca} + \text{I}_2
ightarrow \text{CaI}_2 $? Wait, no, maybe a typo? Wait, the problem says "Ca + ___ → CaI". Wait, maybe it's Ca + I₂ → CaI₂? Wait, the user's image shows "Ca + ___ → CaI". Wait, maybe that's a typo, but assuming it's CaI₂. Wait, no, if it's CaI, then:

Unbalanced: $ \text{Ca} + \text{I}_2
ightarrow \text{CaI} $

Balance I: 2 on left, 1 on right. So put 2 in front of CaI: $ \text{Ca} + \text{I}_2
ightarrow 2\text{CaI} $

Then balance Ca: 2 on right, so 2 in front of Ca: $ 2\text{Ca} + \text{I}_2
ightarrow 2\text{CaI} $

But maybe the intended product is CaI₂. Let's proceed with the given.

Step 1: Identify the reaction type (Problem 8)

Reaction between lead(II) nitrate (Pb(NO₃)₂) and sodium hydroxide (NaOH) to form lead(II) hydroxide (Pb(OH)₂) and sodium nitrate (NaNO₃). Unbalanced equation: $ \text{Pb(NO}_3\text{)}_2 + \text{NaOH}
ightarrow \text{Pb(OH)}_2 + \text{NaNO}_3 $

Step 2: Balance the nitrate (NO₃⁻) ions

On the left, 2 NO₃⁻ (from Pb(NO₃)₂), on the right, 1 NO₃⁻ (from NaNO₃). So put 2 in front of NaNO₃: $ \text{Pb(NO}_3\text{)}_2 + \text{NaOH}
ightarrow \text{Pb(OH)}_2 + 2\text{NaNO}_3 $

Step 3: Balance the sodium (Na⁺) ions

Now, 2 Na on right (from 2 NaNO₃), so put 2 in front of NaOH: $ \text{Pb(NO}_3\text{)}_2 + 2\text{NaOH}
ightarrow \text{Pb(OH)}_2 + 2\text{NaNO}_3 $

Step 4: Check hydroxide (OH⁻)

Left: 2 OH⁻ (from 2 NaOH), right: 2 OH⁻ (from Pb(OH)₂). Pb: 1 on left, 1 on right. So balanced: $ \text{Pb(NO}_3\text{)}_2(aq) + 2\text{NaOH}(aq)
ightarrow \text{Pb(OH)}_2(s) + 2\text{NaNO}_3(aq) $

Step 1: Identify the reaction type (Problem 9)

Combustion of octane (C₈H₁₈) with oxygen (O₂) to form carbon dioxide (CO₂) and water (H₂O). Unbalanced equation: $ \text{C}_8\text{H}_{18}(g) + \text{O}_2(g)
ightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g) $

Step 2: Balance carbon atoms

8 C on left (C₈H₁₈), so put 8 in front of CO₂: $ \text{C}_8\text{H}_{18} + \text{O}_2
ightarrow 8\text{CO}_2 + \text{H}_2\text{O} $

Step 3: Balance hydrogen atoms

18 H on left, so put 9 in front of H₂O (9×2=18 H): $ \text{C}_8\text{H}_{18} + \text{O}_2
ightarrow 8\text{CO}_2 + 9\text{H}_2\text{O} $

Step 4: Balance oxygen atoms

Right: 8×2 + 9×1 = 16 + 9 = 25 O. Left: O₂, so coefficient is 25/2. To eliminate fraction, multiply all coefficients by 2:

$ 2\text{C}_8\text{H}_{18} + 25\text{O}_2
ightarrow 16\text{CO}_2 + 18\text{H}_2\text{O} $

Answers:

Problem 4:

Balanced equation: $ \boldsymbol{2\text{Mg} + \text{O}_2
ightarrow 2\text{MgO}} $

Problem 5:

Balanced equation: $ \boldsymbol{4\text{Fe} + 3\text{O}_2
ightarrow 2\text{Fe}_2\text{O}_3} $

Problem 6:

Reaction: Li + Cl₂ → LiCl (synthesis). Balanced: $ \boldsymbol{2\text{Li} + \text{Cl}_2
ightarrow 2\text{LiCl}} $

Problem 7 (assuming CaI₂, but if CaI):

Balanced: $ \boldsymbol{2\text{Ca} + \text{I}_2
ightarrow 2\text{CaI}} $ (if product is CaI) or $ \boldsymbol{\text{Ca} + \text{I}_2
ightarrow \text{CaI}_2} $ (if product is CaI₂, already balanced)

Problem 8:

Balanced equation: $ \boldsymbol{\text{Pb(NO}_3\text{)}_2(aq) + 2\text{NaOH}(aq)
ightarrow \text{Pb(OH)}_2(s) + 2\text{NaNO}_3(aq)} $

Problem 9:

Balanced equation: $ \boldsymbol{2\text{C}_8\text{H}_{18}(g) + 25\text{O}_2(g)
ightarrow 16\text{CO}_2(g) + 18\text{H}_2\text{O}(g)} $

Answer:

s:

Problem 4:

Balanced equation: $ \boldsymbol{2\text{Mg} + \text{O}_2
ightarrow 2\text{MgO}} $

Problem 5:

Balanced equation: $ \boldsymbol{4\text{Fe} + 3\text{O}_2
ightarrow 2\text{Fe}_2\text{O}_3} $

Problem 6:

Reaction: Li + Cl₂ → LiCl (synthesis). Balanced: $ \boldsymbol{2\text{Li} + \text{Cl}_2
ightarrow 2\text{LiCl}} $

Problem 7 (assuming CaI₂, but if CaI):

Balanced: $ \boldsymbol{2\text{Ca} + \text{I}_2
ightarrow 2\text{CaI}} $ (if product is CaI) or $ \boldsymbol{\text{Ca} + \text{I}_2
ightarrow \text{CaI}_2} $ (if product is CaI₂, already balanced)

Problem 8:

Balanced equation: $ \boldsymbol{\text{Pb(NO}_3\text{)}_2(aq) + 2\text{NaOH}(aq)
ightarrow \text{Pb(OH)}_2(s) + 2\text{NaNO}_3(aq)} $

Problem 9:

Balanced equation: $ \boldsymbol{2\text{C}_8\text{H}_{18}(g) + 25\text{O}_2(g)
ightarrow 16\text{CO}_2(g) + 18\text{H}_2\text{O}(g)} $