key vocabulary
reactants, products, coefficients, subscripts, balanced equation, chemical reaction
day 3: balancing equations (basic)
notes : law of conservation of mass: atoms must be equal on both sides.
practice (day 3 – level 1)
balance the equations:
20. ___ na + ___ cl₂ → ___ nacl
21. ___ mg + ___ o₂ → ___ mgo
22. ___ h₂ + ___ o₂ → ___ h₂o
23. ___ al + ___ o₂ → ___ al₂o₃
24. ___ c₂h₆ + ___ o₂ → ___ co₂ + ___ h₂o
25. ___ fe + ___ h₂o → ___ fe₃o₄ + ___ h₂
26. ___ pb(no₃)₂ → ___ pbo + ___ no₂ + ___ o₂
27. ___ ca(oh)₂ + ___ h₃po₄ → ___ ca₃(po₄)₂ + ___ h₂o

Question

Accepted Answer

### Problem 20:
# Explanation:
## Step1: Balance Cl atoms.
On the left, we have $ \text{Cl}_2 $ (2 Cl atoms), and on the right, $ \text{NaCl} $ has 1 Cl atom. So we need 2 $ \text{NaCl} $ to balance Cl: $ \text{Na} + \text{Cl}_2
ightarrow 2\text{NaCl} $
## Step2: Balance Na atoms.
Now, on the right, we have 2 Na atoms (from 2 $ \text{NaCl} $), so we need 2 Na on the left: $ 2\text{Na} + \text{Cl}_2
ightarrow 2\text{NaCl} $
# Answer:
$ 2 $ $ \text{Na} + 1 $ $ \text{Cl}_2
ightarrow 2 $ $ \text{NaCl} $ (Coefficients: 2, 1, 2)

### Problem 21:
# Explanation:
## Step1: Balance O atoms.
On the left, $ \text{O}_2 $ has 2 O atoms, and on the right, $ \text{MgO} $ has 1 O atom. So we need 2 $ \text{MgO} $ to balance O: $ \text{Mg} + \text{O}_2
ightarrow 2\text{MgO} $
## Step2: Balance Mg atoms.
Now, on the right, we have 2 Mg atoms (from 2 $ \text{MgO} $), so we need 2 Mg on the left: $ 2\text{Mg} + \text{O}_2
ightarrow 2\text{MgO} $
# Answer:
$ 2 $ $ \text{Mg} + 1 $ $ \text{O}_2
ightarrow 2 $ $ \text{MgO} $ (Coefficients: 2, 1, 2)

### Problem 22:
# Explanation:
## Step1: Balance O atoms.
On the left, $ \text{O}_2 $ has 2 O atoms, and on the right, $ \text{H}_2\text{O} $ has 1 O atom. So we need 2 $ \text{H}_2\text{O} $ to balance O: $ \text{H}_2 + \text{O}_2
ightarrow 2\text{H}_2\text{O} $
## Step2: Balance H atoms.
Now, on the right, we have 4 H atoms (from 2 $ \text{H}_2\text{O} $), so we need 2 $ \text{H}_2 $ on the left (since each $ \text{H}_2 $ has 2 H atoms): $ 2\text{H}_2 + \text{O}_2
ightarrow 2\text{H}_2\text{O} $
# Answer:
$ 2 $ $ \text{H}_2 + 1 $ $ \text{O}_2
ightarrow 2 $ $ \text{H}_2\text{O} $ (Coefficients: 2, 1, 2)

### Problem 23:
# Explanation:
## Step1: Balance O atoms.
On the left, $ \text{O}_2 $ has 2 O atoms, and on the right, $ \text{Al}_2\text{O}_3 $ has 3 O atoms. The least common multiple of 2 and 3 is 6. So we need 3 $ \text{O}_2 $ (to get 6 O atoms) and 2 $ \text{Al}_2\text{O}_3 $ (to get 6 O atoms): $ \text{Al} + 3\text{O}_2
ightarrow 2\text{Al}_2\text{O}_3 $
## Step2: Balance Al atoms.
Now, on the right, we have 4 Al atoms (from 2 $ \text{Al}_2\text{O}_3 $), so we need 4 Al on the left: $ 4\text{Al} + 3\text{O}_2
ightarrow 2\text{Al}_2\text{O}_3 $
# Answer:
$ 4 $ $ \text{Al} + 3 $ $ \text{O}_2
ightarrow 2 $ $ \text{Al}_2\text{O}_3 $ (Coefficients: 4, 3, 2)

### Problem 24:
# Explanation:
## Step1: Balance C atoms.
On the left, $ \text{C}_2\text{H}_6 $ has 2 C atoms, so we need 2 $ \text{CO}_2 $ on the right: $ \text{C}_2\text{H}_6 + \text{O}_2
ightarrow 2\text{CO}_2 + \text{H}_2\text{O} $
## Step2: Balance H atoms.
On the left, $ \text{C}_2\text{H}_6 $ has 6 H atoms, so we need 3 $ \text{H}_2\text{O} $ (since each $ \text{H}_2\text{O} $ has 2 H atoms: $ 6 \div 2 = 3 $): $ \text{C}_2\text{H}_6 + \text{O}_2
ightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} $
## Step3: Balance O atoms.
On the right, we have $ 2 \times 2 + 3 \times 1 = 7 $ O atoms. But $ \text{O}_2 $ is diatomic, so we need $ \frac{7}{2} $ $ \text{O}_2 $, but we use whole numbers. Multiply all coefficients by 2 to eliminate the fraction: $ 2\text{C}_2\text{H}_6 + 7\text{O}_2
ightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} $
# Answer:
$ 2 $ $ \text{C}_2\text{H}_6 + 7 $ $ \text{O}_2
ightarrow 4 $ $ \text{CO}_2 + 6 $ $ \text{H}_2\text{O} $ (Coefficients: 2, 7, 4, 6)

### Problem 25:
# Explanation:
## Step1: Balance Fe atoms.
On the right, $ \text{Fe}_3\text{O}_4 $ has 3 Fe atoms, so we need 3 Fe on the left: $ 3\text{Fe} + \text{H}_2\text{O}
ightarrow \text{Fe}_3\text{O}_4 + \text{H}_2 $
## Step2: Balance O atoms.
On the right, $ \text{Fe}_3\text{O}_4 $ has 4 O atoms, so we need 4 $ \text{H}_2\text{O} $ on the left: $ 3\text{Fe} + 4\text{H}_2\text{O}
ightarrow \text{Fe}_3\text{O}_4 + \text{H}_2 $
## Step3: Balance H atoms.
On the left, $ 4\text{H}_2\text{O} $ has 8 H atoms, so we need 4 $ \text{H}_2 $ on the right: $ 3\text{Fe} + 4\text{H}_2\text{O}
ightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2 $
# Answer:
$ 3 $ $ \text{Fe} + 4 $ $ \text{H}_2\text{O}
ightarrow 1 $ $ \text{Fe}_3\text{O}_4 + 4 $ $ \text{H}_2 $ (Coefficients: 3, 4, 1, 4)

### Problem 26:
# Explanation:
## Step1: Balance Pb atoms.
On the left, $ \text{Pb(NO}_3\text{)}_2 $ has 1 Pb atom, and on the right, $ \text{PbO} $ has 1 Pb atom. So Pb is balanced for now.
## Step2: Balance N atoms.
On the left, $ \text{Pb(NO}_3\text{)}_2 $ has 2 N atoms, so we need 2 $ \text{NO}_2 $ on the right: $ \text{Pb(NO}_3\text{)}_2
ightarrow \text{PbO} + 2\text{NO}_2 + \text{O}_2 $
## Step3: Balance O atoms.
On the left, $ \text{Pb(NO}_3\text{)}_2 $ has $ 2 \times 3 = 6 $ O atoms. On the right, $ \text{PbO} $ has 1 O, $ 2\text{NO}_2 $ has $ 2 \times 2 = 4 $ O, so total O on right is $ 1 + 4 + 2x $ (where $ x $ is the coefficient of $ \text{O}_2 $). So $ 6 = 1 + 4 + 2x
ightarrow 2x = 1
ightarrow x = \frac{1}{2} $. Multiply all coefficients by 2: $ 2\text{Pb(NO}_3\text{)}_2
ightarrow 2\text{PbO} + 4\text{NO}_2 + \text{O}_2 $
# Answer:
$ 2 $ $ \text{Pb(NO}_3\text{)}_2
ightarrow 2 $ $ \text{PbO} + 4 $ $ \text{NO}_2 + 1 $ $ \text{O}_2 $ (Coefficients: 2, 2, 4, 1)

### Problem 27:
# Explanation:
## Step1: Balance Ca atoms.
On the right, $ \text{Ca}_3(\text{PO}_4)_2 $ has 3 Ca atoms, so we need 3 $ \text{Ca(OH)}_2 $ on the left: $ 3\text{Ca(OH)}_2 + \text{H}_3\text{PO}_4
ightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{O} $
## Step2: Balance P atoms.
On the right, $ \text{Ca}_3(\text{PO}_4)_2 $ has 2 P atoms, so we need 2 $ \text{H}_3\text{PO}_4 $ on the left: $ 3\text{Ca(OH)}_2 + 2\text{H}_3\text{PO}_4
ightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{O} $
## Step3: Balance H and O atoms.
On the left, $ 3\text{Ca(OH)}_2 $ has $ 3 \times 2 = 6 $ H, and $ 2\text{H}_3\text{PO}_4 $ has $ 2 \times 3 = 6 $ H, so total H is $ 6 + 6 = 12 $. On the right, $ \text{H}_2\text{O} $ has 2 H per molecule, so we need $ \frac{12}{2} = 6 $ $ \text{H}_2\text{O} $. Check O: Left has $ 3 \times 2 + 2 \times 4 = 6 + 8 = 14 $. Right has $ 2 \times 4 + 6 \times 1 = 8 + 6 = 14 $. Balanced.
# Answer:
$ 3 $ $ \text{Ca(OH)}_2 + 2 $ $ \text{H}_3\text{PO}_4
ightarrow 1 $ $ \text{Ca}_3(\text{PO}_4)_2 + 6 $ $ \text{H}_2\text{O} $ (Coefficients: 3, 2, 1, 6)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 20:

Step1: Balance Cl atoms.

Step2: Balance Na atoms.

Step1: Balance O atoms.

Step2: Balance Mg atoms.

Step1: Balance O atoms.

Step2: Balance H atoms.

Answer:

Problem 21: