Question

write balanced chemical equations for the following reactions:

1. when sodium hydroxide reacts with sulfuric acid, sodium sulfate and water are formed.
2. when fluorine gas reacts with calcium metal, calcium fluoride powder is created.
3. magnesium catches fire in the presence of oxygen to form magnesium oxide.
4. silver reacts with the sulfur in the air and produces silver sulfide.
5. nitrogen and hydrogen yield ammonia.
6. iron and lead (ii) sulfate react to form iron (ii) sulfate and lead.
7. tin (ii) oxide and hydrogen gas combine to form tin and water.
8. gases methane and oxygen react to form carbon dioxide and water.
9. sodium combines with chlorine gas to form sodium chloride.
10. when potassium chlorate is heated, potassium chloride and oxygen result.

Explanation:

Step1: Analyze Reaction 3 (Magnesium + Oxygen)

Identify reactants (Mg, \(O_2\)) and product (MgO). Unbalanced equation: \(Mg + O_2
ightarrow MgO\).

Step2: Balance Oxygen

Oxygen has 2 atoms on left, 1 on right. Add coefficient 2 to MgO: \(Mg + O_2
ightarrow 2MgO\).

Step3: Balance Magnesium

Now Mg has 2 atoms on right, so add coefficient 2 to Mg: \(2Mg + O_2
ightarrow 2MgO\).

Step1: Analyze Reaction 4 (Silver + Sulfur)

Reactants: Ag, S (from air, \(S_8\) or S, assume S for simplicity), product: \(Ag_2S\). Unbalanced: \(Ag + S
ightarrow Ag_2S\).

Step2: Balance Silver

Ag has 2 atoms on right, add 2 to Ag: \(2Ag + S
ightarrow Ag_2S\). Sulfur is balanced (1 on each side).

Step1: Analyze Reaction 5 (Nitrogen + Hydrogen)

Reactants: \(N_2\), \(H_2\), product: \(NH_3\). Unbalanced: \(N_2 + H_2
ightarrow NH_3\).

Step2: Balance Nitrogen

\(N_2\) has 2 N, so \(NH_3\) needs coefficient 2: \(N_2 + H_2
ightarrow 2NH_3\).

Step3: Balance Hydrogen

\(2NH_3\) has 6 H, so \(H_2\) gets 3: \(N_2 + 3H_2
ightarrow 2NH_3\) (balanced).

Step1: Analyze Reaction 6 (Iron + \(PbSO_4\))

Reactants: Fe, \(PbSO_4\), products: \(FeSO_4\), Pb. Unbalanced: \(Fe + PbSO_4
ightarrow FeSO_4 + Pb\). Check atoms: Fe (1), Pb (1), \(SO_4^{2-}\) (1) on each side. Already balanced: \(Fe + PbSO_4 = FeSO_4 + Pb\).

Step1: Analyze Reaction 7 (Tin(II) Oxide + \(H_2\))

Reactants: \(SnO\) (or \(SnO_2\)? Wait, Tin(II) oxide is \(SnO\)? No, Tin(II) oxide: \(SnO\) (Sn²⁺, O²⁻), but common is \(SnO_2\) for Tin(IV). Wait, problem says Tin(II) oxide: \(SnO\). Reactants: \(SnO\), \(H_2\), products: Sn, \(H_2O\). Unbalanced: \(SnO + H_2
ightarrow Sn + H_2O\). Check O: 1 on left, 1 on right. H: 2 on left, 2 on right. Sn: 1 on left, 1 on right. Wait, balanced? Wait, \(SnO\) (Sn²⁺) + \(H_2\) → Sn + \(H_2O\). Yes, balanced: \(SnO + H_2 = Sn + H_2O\). Wait, maybe typo, if Tin(II) oxide is \(SnO_2\) (Tin(IV)? No, Tin(II) is \(SnO\), Tin(IV) is \(SnO_2\). Assuming \(SnO\), equation is balanced. If \(SnO_2\): \(SnO_2 + 2H_2
ightarrow Sn + 2H_2O\). Let's check problem: "Tin (II) oxide" → \(SnO\), so balanced as \(SnO + H_2
ightarrow Sn + H_2O\).

Step1: Analyze Reaction 8 (Methane + Oxygen)

Reactants: \(CH_4\), \(O_2\), products: \(CO_2\), \(H_2O\). Unbalanced: \(CH_4 + O_2
ightarrow CO_2 + H_2O\).

Step2: Balance Carbon

C: 1 on left, 1 on right (good).

Step3: Balance Hydrogen

\(CH_4\) has 4 H, so \(H_2O\) gets 2: \(CH_4 + O_2
ightarrow CO_2 + 2H_2O\).

Step4: Balance Oxygen

\(CO_2\) has 2 O, \(2H_2O\) has 2 O, total 4 O on right. \(O_2\) has 2 O per molecule, so add 2: \(CH_4 + 2O_2
ightarrow CO_2 + 2H_2O\) (balanced).

Step1: Analyze Reaction 9 (Sodium + Chlorine)

Reactants: Na, \(Cl_2\), product: NaCl. Unbalanced: \(Na + Cl_2
ightarrow NaCl\).

Step2: Balance Chlorine

\(Cl_2\) has 2 Cl, so NaCl gets 2: \(Na + Cl_2
ightarrow 2NaCl\).

Step3: Balance Sodium

Add 2 to Na: \(2Na + Cl_2
ightarrow 2NaCl\) (balanced).

Step1: Analyze Reaction 10 (Potassium Chlorate Decomposition)

Reactant: \(KClO_3\), products: KCl, \(O_2\). Unbalanced: \(KClO_3
ightarrow KCl + O_2\).

Step2: Balance Oxygen

\(KClO_3\) has 3 O, \(O_2\) has 2. Find LCM of 3 and 2 (6). So \(KClO_3\) coefficient 2, \(O_2\) coefficient 3: \(2KClO_3
ightarrow KCl + 3O_2\).

Step3: Balance Potassium and Chlorine

K and Cl: 2 on left, so KCl gets 2: \(2KClO_3
ightarrow 2KCl + 3O_2\) (balanced).

Answer:

1. \(2Mg + O_2

ightarrow 2MgO\)

1. \(2Ag + S

ightarrow Ag_2S\) (assuming S is atomic; if \(S_8\): \(16Ag + S_8
ightarrow 8Ag_2S\))

1. \(N_2 + 3H_2

ightarrow 2NH_3\)

1. \(Fe + PbSO_4

ightarrow FeSO_4 + Pb\)

1. \(SnO + H_2

ightarrow Sn + H_2O\) (or \(SnO_2 + 2H_2
ightarrow Sn + 2H_2O\) if Tin(IV) oxide)

1. \(CH_4 + 2O_2

ightarrow CO_2 + 2H_2O\)

1. \(2Na + Cl_2

ightarrow 2NaCl\)

1. \(2KClO_3 \xrightarrow{\Delta} 2KCl + 3O_2\uparrow\)