Question

balancing equations
balance the following chemical equations.

1. \t__ ch₄ + o₂ → co₂ + __ h₂o
2. \t__ na⁺ + cl⁻ → __ nacl
3. \t__ al + o₂ → __ al₂o₃
4. \t__ n₂ + h₂ → __ nh₃
5. \t__ co(g) + h₂(g) → c₈h₁₈(l) + __ h₂o
6. \t__ fe₂o₃(s) + co(g) → fe(l) + __ co₂(g)
7. \t__ h₂so₄ + pb(oh)₄ → pb(so₄)₂ + __ h₂o
8. \t__ al + hcl → alcl₃ + __ h₂
9. \t__ ca₃(po₄)₂ + h₂so₄ → caso₄ + __ ca(h₂po₄)₂

__ h₃po₄ + hcl → pcl₅ + __ h₂o

Explanation:

Step1: Balance \( \boldsymbol{CH_4 + O_2

ightarrow CO_2 + H_2O} \)
- **C**: 1 on left (\(CH_4\)), 1 on right (\(CO_2\)) – balanced.
- **H**: 4 on left (\(CH_4\)), 2 on right (\(H_2O\)) → set \(H_2O\) coefficient to 2 (now 4 H).
- **O**: 2 on left (\(O_2\)), \(2 + 2 = 4\) on right (\(CO_2 + 2H_2O\)) → set \(O_2\) coefficient to 2.
Equation: \(1\ CH_4 + 2\ O_2 ightarrow 1\ CO_2 + 2\ H_2O\) (Wait, original \(H_2O\) had a 3? Correct: \(CH_4 + 2O_2 ightarrow CO_2 + 2H_2O\) (H: 4→4, O: 4→4).

Step2: Balance \( \boldsymbol{Na^+ + Cl^-

ightarrow NaCl} \)
- \(Na^+\) and \(Cl^-\) combine 1:1. Coefficients: \(1\ Na^+ + 1\ Cl^- ightarrow 1\ NaCl\).

Step3: Balance \( \boldsymbol{Al + O_2

ightarrow Al_2O_3} \)
- **Al**: 1 on left, 2 on right → set \(Al\) to 2.
- **O**: 2 on left, 3 on right → LCM of 2,3 is 6. Set \(O_2\) to 3 (6 O), \(Al_2O_3\) to 2 (6 O).
Equation: \(4\ Al + 3\ O_2 ightarrow 2\ Al_2O_3\) (Al: 4→4, O: 6→6).

Step4: Balance \( \boldsymbol{N_2 + H_2

ightarrow NH_3} \)
- **N**: 2 on left, 1 on right → set \(NH_3\) to 2 (2 N).
- **H**: 2 on left, 6 on right (\(2NH_3\)) → set \(H_2\) to 3 (6 H).
Equation: \(1\ N_2 + 3\ H_2 ightarrow 2\ NH_3\).

Step5: Balance \( \boldsymbol{CO + H_2

ightarrow C_8H_{18} + H_2O} \)
- **C**: 1 on left (\(CO\)), 8 on right (\(C_8H_{18}\)) → set \(CO\) to 8 (8 C).
- **H**: 2 on left (\(H_2\)), \(18 + 2x\) on right (\(C_8H_{18} + xH_2O\)). Wait, better: \(8CO + H_2 ightarrow C_8H_{18} + H_2O\).
- **O**: 8 on left (\(8CO\)), \(x\) on right (\(H_2O\)) → \(x=8\) (8 O).
- **H**: \(2y = 18 + 8(2)\)? No, correct: \(8CO + 17H_2 ightarrow 1C_8H_{18} + 8H_2O\) (C:8→8, O:8→8, H:34→18+16=34).

Step6: Balance \( \boldsymbol{Fe_2O_3 + CO

ightarrow Fe + CO_2} \)
- **Fe**: 2 on left (\(Fe_2O_3\)), 1 on right → set \(Fe\) to 2.
- **O**: \(3 + x = 2y\) (Fe₂O₃ + xCO → 2Fe + yCO₂). Each CO → CO₂ gains 1 O. Fe₂O₃ has 3 O to donate → x=3, y=3.
Equation: \(1\ Fe_2O_3 + 3\ CO ightarrow 2\ Fe + 3\ CO_2\) (Fe:2→2, O:6→6, C:3→3).

Step7: Balance \( \boldsymbol{H_2SO_4 + Pb(OH)_4

ightarrow Pb(SO_4)_2 + H_2O} \)
- **SO₄²⁻**: 1 on left, 2 on right → set \(H_2SO_4\) to 2 (2 SO₄²⁻).
- **Pb**: 1 on left (\(Pb(OH)_4\)), 1 on right (\(Pb(SO_4)_2\)) – balanced.
- **H**: \(2(2) + 4 = 8\) on left, 2 on right (\(H_2O\)) → set \(H_2O\) to 4 (8 H).
- **O**: \(2(4) + 4 = 12\) on left, \(2(4) + 4 = 12\) on right – balanced.
Equation: \(2\ H_2SO_4 + 1\ Pb(OH)_4 ightarrow 1\ Pb(SO_4)_2 + 4\ H_2O\).

Step8: Balance \( \boldsymbol{Al + HCl

ightarrow AlCl_3 + H_2} \)
- **Al**: 1 on left, 1 on right → set \(Al\) to 1.
- **Cl**: 3 on right (\(AlCl_3\)), 1 on left → set \(HCl\) to 3 (3 Cl).
- **H**: 3 on left (\(3HCl\)), 2 on right (\(H_2\)) → LCM 6. Set \(HCl\) to 6, \(H_2\) to 3, \(Al\) to 2, \(AlCl_3\) to 2.
Equation: \(2\ Al + 6\ HCl ightarrow 2\ AlCl_3 + 3\ H_2\) (Al:2→2, Cl:6→6, H:6→6).

Step9: Balance \( \boldsymbol{Ca_3(PO_4)_2 + H_2SO_4

ightarrow CaSO_4 + Ca(H_2PO_4)_2} \)
- **Ca**: 3 on left, \(1 + 1 = 2\) on right? Wait, \(Ca(H_2PO_4)_2\) has 1 Ca, \(CaSO_4\) has 1 Ca. Let \(CaSO_4 = x\), \(Ca(H_2PO_4)_2 = 1\) (2 PO₄³⁻).
- **PO₄³⁻**: 2 on left (\(Ca_3(PO_4)_2\)), 2 on right (\(Ca(H_2PO_4)_2\)) – balanced.
- **Ca**: 3 on left, \(x + 1 = 3\) → \(x=2\) (2 \(CaSO_4\)).
- **SO₄²⁻**: \(x = 2\) on right, so \(H_2SO_4 = 2\) (2 SO₄²⁻).
- **H**: \(2(2) = 4\) on left (\(2H_2SO_4\)), \(2(2) = 4\) on right (\(Ca(H_2PO_4)_2\)) – balanced.
Equation: \(1\ Ca_3(PO_4)_2 + 2\ H_2SO_4 ightarrow 2\ CaSO_4 + 1\ Ca(H_2PO_4)_2\).

Step10: Balance \( \boldsymbol{H_3PO_4 + HCl

ightarrow PCl_5 + H_2O} \)
- **P**: 1 on left, 1 on right – balanced.
- **Cl**: 1 on left, 5 on right → set \(HCl\) to 5 (5 Cl).
- **H**: \(3 + 5 = 8\) on left, 2 on right (\(H_2O\)) → set \(H_2O\) to 4 (8 H).
- **O**: 4 on left (\(H_3PO_4\)), 4 on right (\(4H_2O\)) – balanced.
Equation: \(1\ H_3PO_4 + 5\ HCl ightarrow 1\ PCl_5 + 4\ H_2O\).

Answers (Coefficients):

1. \( \boldsymbol{1\ CH_4 + 2\ O_2 ightarrow 1\ CO_2 + 2\ H_2O} \) (Correcting the 3 to 2 for H balance)
2. \( \boldsymbol{1\ Na^+ + 1\ Cl^- ightarrow 1\ NaCl} \)
3. \( \boldsymbol{4\ Al + 3\ O_2 ightarrow 2\ Al_2O_3} \)
4. \( \boldsymbol{1\ N_2 + 3\ H_2 ightarrow 2\ NH_3} \)
5. \( \boldsymbol{8\ CO + 17\ H_2 ightarrow 1\ C_8H_{18} + 8\ H_2O} \)
6. \( \boldsymbol{1\ Fe_2O_3 + 3\ CO ightarrow 2\ Fe + 3\ CO_2} \)
7. \( \boldsymbol{2\ H_2SO_4 + 1\ Pb(OH)_4 ightarrow 1\ Pb(SO_4)_2 + 4\ H_2O} \)
8. \( \boldsymbol{2\ Al + 6\ HCl ightarrow 2\ AlCl_3 + 3\ H_2} \)
9. \( \boldsymbol{1\ Ca_3(PO_4)_2 + 2\ H_2SO_4 ightarrow 2\ CaSO_4 + 1\ Ca(H_2PO_4)_2} \)
10. \( \boldsymbol{1\ H_3PO_4 + 5\ HCl ightarrow 1\ PCl_5 + 4\ H_2O} \)

(Note: For problem 1, the initial 3 for \(H_2O\) was incorrect; correct is 2 to balance H and O.)

Answer:

s (Coefficients):

1. \( \boldsymbol{1\ CH_4 + 2\ O_2

ightarrow 1\ CO_2 + 2\ H_2O} \) (Correcting the 3 to 2 for H balance)

1. \( \boldsymbol{1\ Na^+ + 1\ Cl^-

ightarrow 1\ NaCl} \)

1. \( \boldsymbol{4\ Al + 3\ O_2

ightarrow 2\ Al_2O_3} \)

1. \( \boldsymbol{1\ N_2 + 3\ H_2

ightarrow 2\ NH_3} \)

1. \( \boldsymbol{8\ CO + 17\ H_2

ightarrow 1\ C_8H_{18} + 8\ H_2O} \)

1. \( \boldsymbol{1\ Fe_2O_3 + 3\ CO

ightarrow 2\ Fe + 3\ CO_2} \)

1. \( \boldsymbol{2\ H_2SO_4 + 1\ Pb(OH)_4

ightarrow 1\ Pb(SO_4)_2 + 4\ H_2O} \)

1. \( \boldsymbol{2\ Al + 6\ HCl

ightarrow 2\ AlCl_3 + 3\ H_2} \)

1. \( \boldsymbol{1\ Ca_3(PO_4)_2 + 2\ H_2SO_4

ightarrow 2\ CaSO_4 + 1\ Ca(H_2PO_4)_2} \)

1. \( \boldsymbol{1\ H_3PO_4 + 5\ HCl

ightarrow 1\ PCl_5 + 4\ H_2O} \)

(Note: For problem 1, the initial 3 for \(H_2O\) was incorrect; correct is 2 to balance H and O.)