2) \underline{6} k + \underline{\quad} b₂o₃ → \underline{3} k₂o + \underline{2} b\
3) \underline{\quad} hcl + \underline{\quad} naoh → \underline{\quad} nacl + \underline{\quad} h₂o\
4) \underline{\quad} na + \underline{2} nano₃ → \underline{2} na₂o + \underline{4} n₂\
5) \underline{\quad} c + \underline{\quad} s₈ → \underline{\quad} cs₂\
6) \underline{\quad} na + \underline{\quad} o₂ → \underline{\quad} na₂o\
7) \underline{\quad} n₂ + \underline{\quad} o₂ → \underline{10} n₂o₅

Question

Accepted Answer

### Problem 2:
# Explanation:
## Step1: Balance O atoms.
On the left, $ B_2O_3 $ has 3 O. On the right, $ K_2O $ has 1 O. So put 3 in front of $ K_2O $ to make O: $ 3 \times 1 = 3 $. Now equation: $ \underline{\quad} K + \underline{\quad} B_2O_3
ightarrow \underline{3} K_2O + \underline{\quad} B $
## Step2: Balance K atoms.
$ 3 K_2O $ has $ 3 \times 2 = 6 $ K. So put 6 in front of K. Equation: $ \underline{6} K + \underline{\quad} B_2O_3
ightarrow \underline{3} K_2O + \underline{\quad} B $
## Step3: Balance B atoms.
$ B_2O_3 $ has 2 B. So put 2 in front of B. And check $ B_2O_3 $: only 1 needed. Final equation: $ \underline{6} K + \underline{1} B_2O_3
ightarrow \underline{3} K_2O + \underline{2} B $
# Answer: 6, 1, 3, 2

### Problem 3:
# Explanation:
## Step1: This is a neutralization reaction. HCl + NaOH → NaCl + H₂O. Check atoms:
H: 1 (HCl) + 1 (NaOH) = 2 (H₂O: 2 H). Cl: 1 (HCl) = 1 (NaCl). Na: 1 (NaOH) = 1 (NaCl). O: 1 (NaOH) = 1 (H₂O). So all atoms are balanced with 1 in each blank.
# Answer: 1, 1, 1, 1

### Problem 4:
# Explanation:
## Step1: Balance N atoms.
Left: $ NaNO_3 $ has 1 N. Right: $ N_2 $ has 2 N. So put 2 in front of $ NaNO_3 $ to make N: $ 2 \times 1 = 2 $, matching $ N_2 $. Now equation: $ \underline{\quad} Na + \underline{2} NaNO_3
ightarrow \underline{\quad} Na_2O + \underline{\quad} N_2 $
## Step2: Balance O atoms.
$ 2 NaNO_3 $ has $ 2 \times 3 = 6 $ O. Right: $ Na_2O $ has 1 O. So put 6 in front of $ Na_2O $ (since $ 6 \times 1 = 6 $ O). Equation: $ \underline{\quad} Na + \underline{2} NaNO_3
ightarrow \underline{6} Na_2O + \underline{\quad} N_2 $
## Step3: Balance Na atoms.
Left: Na (let's say x) + $ 2 NaNO_3 $ (2 Na) = x + 2. Right: $ 6 Na_2O $ (12 Na). So x + 2 = 12 → x = 10. Put 10 in front of Na. Equation: $ \underline{10} Na + \underline{2} NaNO_3
ightarrow \underline{6} Na_2O + \underline{1} N_2 $ (Check N: 2, O: 6, Na: 10 + 2 = 12, and $ 6 Na_2O $ has 12 Na. Balanced.)
# Answer: 10, 2, 6, 1

### Problem 5:
# Explanation:
## Step1: Balance S atoms.
Left: $ S_8 $ has 8 S. Right: $ CS_2 $ has 2 S. So number of $ CS_2 $ needed: $ 8 \div 2 = 4 $. Put 4 in front of $ CS_2 $. Equation: $ \underline{\quad} C + \underline{\quad} S_8
ightarrow \underline{4} CS_2 $
## Step2: Balance C atoms.
$ 4 CS_2 $ has 4 C. So put 4 in front of C. Equation: $ \underline{4} C + \underline{\quad} S_8
ightarrow \underline{4} CS_2 $
## Step3: Balance S atoms.
$ 4 CS_2 $ has $ 4 \times 2 = 8 $ S, which matches $ S_8 $ (1 $ S_8 $ has 8 S). So put 1 in front of $ S_8 $. Final equation: $ \underline{4} C + \underline{1} S_8
ightarrow \underline{4} CS_2 $
# Answer: 4, 1, 4

### Problem 6:
# Explanation:
## Step1: Balance O atoms.
Left: $ O_2 $ has 2 O. Right: $ Na_2O $ has 1 O. So put 2 in front of $ Na_2O $ (2 O). Equation: $ \underline{\quad} Na + \underline{\quad} O_2
ightarrow \underline{2} Na_2O $
## Step2: Balance Na atoms.
$ 2 Na_2O $ has $ 2 \times 2 = 4 $ Na. So put 4 in front of Na. Equation: $ \underline{4} Na + \underline{\quad} O_2
ightarrow \underline{2} Na_2O $
## Step3: Balance O atoms.
$ 2 Na_2O $ has 2 O, and $ O_2 $ has 2 O. So put 1 in front of $ O_2 $. Final equation: $ \underline{4} Na + \underline{1} O_2
ightarrow \underline{2} Na_2O $
# Answer: 4, 1, 2

### Problem 7:
# Explanation:
## Step1: Balance N atoms.
Left: $ N_2 $ has 2 N. Right: $ N_2O_5 $ has 2 N. Let's say we have x $ N_2 $ and y $ N_2O_5 $. But to balance O:
Left: $ O_2 $ has 2 O. Right: $ N_2O_5 $ has 5 O. Let's put 2 in front of $ N_2O_5 $ (10 O) and 5 in front of $ O_2 $ (10 O). Then N: $ 2 N_2O_5 $ has 4 N, so put 2 in front of $ N_2 $ (4 N). Equation: $ \underline{2} N_2 + \underline{5} O_2
ightarrow \underline{2} N_2O_5 $
# Answer: 2, 5, 2

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QUESTION IMAGE

Question

Explanation:

Problem 2:

Step1: Balance O atoms.

Step2: Balance K atoms.

Step3: Balance B atoms.

Step1: This is a neutralization reaction. HCl + NaOH → NaCl + H₂O. Check atoms:

Step1: Balance N atoms.

Step2: Balance O atoms.

Step3: Balance Na atoms.

Answer:

Problem 3: