Question

1. __ kno₃ + _ c₁₂h₂₂o₁₁ → __ n₂ + co₂ + h₂o + __ k₂co₃

Explanation:

Step1: Identify elements

We have K, N, O, C, H in reactants and products. Let's denote coefficients as \( a \) for \( \text{KNO}_3 \), \( b \) for \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \), \( c \) for \( \text{N}_2 \), \( d \) for \( \text{CO}_2 \), \( e \) for \( \text{H}_2\text{O} \), \( f \) for \( \text{K}_2\text{CO}_3 \).

Step2: Balance K

From K: \( a = 2f \) (since \( \text{KNO}_3 \) has 1 K, \( \text{K}_2\text{CO}_3 \) has 2 K)

Step3: Balance N

From N: \( a = 2c \) ( \( \text{KNO}_3 \) has 1 N, \( \text{N}_2 \) has 2 N)

Step4: Balance C

From C: \( 12b = d + f \) ( \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \) has 12 C, \( \text{CO}_2 \) has 1 C, \( \text{K}_2\text{CO}_3 \) has 1 C)

Step5: Balance H

From H: \( 22b = 2e \) ⇒ \( e = 11b \)

Step6: Balance O

From O: \( 3a + 11b = 2d + e + 3f \)

Let's assume \( b = 1 \) (simplest for sugar). Then:

- From K: \( a = 2f \)
- From N: \( a = 2c \)
- From C: \( 12 = d + f \)
- From H: \( e = 11 \)
- From O: \( 3a + 11 = 2d + 11 + 3f \) ⇒ \( 3a = 2d + 3f \)

Substitute \( d = 12 - f \) and \( a = 2f \) into O balance:

\( 3(2f) = 2(12 - f) + 3f \)

\( 6f = 24 - 2f + 3f \)

\( 6f = 24 + f \)

\( 5f = 24 \)? Wait, no, maybe \( b = 1 \) is not good. Wait, let's try redox approach.

\( \text{KNO}_3 \) is oxidizing agent (N from +5 to 0), \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \) is reducing agent (C from 0 to +4, H from +1 to +1, O from -2 to -2? Wait, C in sugar: formula \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \), oxidation state of C: let x be C's oxidation state. \( 12x + 22(+1) + 11(-2) = 0 \) ⇒ \( 12x + 22 - 22 = 0 \) ⇒ \( x = 0 \). In \( \text{CO}_2 \), C is +4, in \( \text{K}_2\text{CO}_3 \), C is +4. So each C atom loses 4 electrons. There are 12 C atoms, so total electrons lost: \( 12 \times 4 = 48 \).

In \( \text{KNO}_3 \), N is +5, in \( \text{N}_2 \) is 0, so each N gains 5 electrons. There are 2 N atoms per \( \text{N}_2 \), so per \( \text{KNO}_3 \), 1 N gains 5, so per \( \text{N}_2 \), 2 N gain 10 electrons.

Let number of \( \text{N}_2 \) be \( c \), so electrons gained: \( 10c \). Electrons lost: 48 (from 1 \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)). So \( 10c = 48 \)? No, wait, \( b \) is number of sugar molecules. So electrons lost per \( b \) sugar: \( 12b \times 4 = 48b \). Electrons gained per \( a \) \( \text{KNO}_3 \): each \( \text{KNO}_3 \) has 1 N, so electrons gained per \( \text{KNO}_3 \): 5, so total electrons gained: \( 5a \).

Thus, \( 5a = 48b \). Let's find smallest integers: \( a = 48 \), \( b = 5 \)? Wait, no, maybe I made a mistake. Wait, let's try balancing by trial.

Let's start with \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \) (sucrose) as 1. Then C: 12, so \( d + f = 12 \). H: 22, so \( e = 11 \). K: from \( \text{KNO}_3 \), \( a = 2f \). N: \( a = 2c \). O: \( 3a + 11 = 2d + 11 + 3f \) ⇒ \( 3a = 2d + 3f \). But \( d = 12 - f \), so \( 3a = 2(12 - f) + 3f = 24 + f \). Also, \( a = 2f \), so \( 3(2f) = 24 + f \) ⇒ \( 6f = 24 + f \) ⇒ \( 5f = 24 \) ⇒ \( f = 24/5 \), not integer. So \( b \) must be 5? Wait, no, let's try \( b = 1 \), but adjust. Wait, maybe the reaction is:

\( \text{KNO}_3 + \text{C}_{12}\text{H}_{22}\text{O}_{11} ightarrow \text{N}_2 + \text{CO}_2 + \text{H}_2\text{O} + \text{K}_2\text{CO}_3 \)

Let's use oxidation states properly.

Oxidation of C: in \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \), C is 0 (as calculated), in \( \text{CO}_2 \) and \( \text{K}_2\text{CO}_3 \), C is +4. So each C loses 4 e⁻, 12 C lose 48 e⁻ per sucrose.

Reduction of N: in \( \text{KNO}_3 \), N is +5, in \( \text{N}_2 \), 0. Each N gains 5 e⁻, so 2 N (per \( \text{N}_2 \)) gain 10 e⁻.

So moles of \( \text{KNO}_3 \) (a) and sucrose (b) must satisfy \( 5a = 48b \). The smallest integers: a=48, b=5. Let's check:

a=48, b=5.

K: a=48, so f=24 (since a=2f ⇒ f=24).

C: 12b=60, so d + f=60 ⇒ d=60 -24=36.

H: 22b=110, so e=55 (since e=11b=55).

N: a=48, so c=24 (since a=2c ⇒ c=24).

Now check O:

Left: 3a + 11b = 3*48 + 11*5 = 144 + 55 = 199.

Right: 2d + e + 3f = 2*36 + 55 + 3*24 = 72 + 55 + 72 = 199. Perfect!

So coefficients:

\( \text{KNO}_3 \): 48, \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \): 5, \( \text{N}_2 \): 24, \( \text{CO}_2 \): 36, \( \text{H}_2\text{O} \): 55, \( \text{K}_2\text{CO}_3 \): 24.

Wait, but let's verify:

Reactants:
- K: 48 (from 48 KNO3)
- N: 48 (from 48 KNO3)
- O: 3*48 + 11*5 = 144 + 55 = 199
- C: 12*5 = 60
- H: 22*5 = 110

Products:
- N2: 24*2 = 48 N (correct)
- CO2: 36*1 = 36 C; K2CO3:24*1=24 C; total C: 36+24=60 (correct)
- H2O:55*2=110 H (correct)
- K2CO3:24*2=48 K (correct)
- O: CO2:36*2=72; H2O:55*1=55; K2CO3:24*3=72; total O:72+55+72=199 (correct)

Yes, that works.

Answer:

48 \( \text{KNO}_3 \) + 5 \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \) → 24 \( \text{N}_2 \) + 36 \( \text{CO}_2 \) + 55 \( \text{H}_2\text{O} \) + 24 \( \text{K}_2\text{CO}_3 \)

So the coefficients are 48, 5, 24, 36, 55, 24.