A mixture of `Li_(2)CO_(3)` and `Na_(2)CO_(3)` is heated strongly in an open vessel .If the loss in the mass of mixture is `(220)/(9)` % .The molar ratio of `Li_(2)CO_(3)` and `Na_(2)CO_(3)` in the initial mixture is (Li = 7,Na = 23)

Options

A. `1:1`
B. `1:2`
C. `2:1`
D. `11:4`

Answer

To solve the problem, we need to find the molar ratio of lithium carbonate (Li₂CO₃) and sodium carbonate (Na₂CO₃) in a mixture after heating. Here’s a step-by-step breakdown of the solution:

### Step 1: Define Variables
Let:
- \( X \) = moles of Li₂CO₃
- \( Y \) = moles of Na₂CO₃

### Step 2: Molar Masses
The molar masses of the compounds are:
- Molar mass of Li₂CO₃ = 2(7) + 12 + 3(16) = 74 g/mol
- Molar mass of Na₂CO₃ = 2(23) + 12 + 3(16) = 106 g/mol
- Molar mass of CO₂ = 12 + 2(16) = 44 g/mol

### Step 3: Reaction of Li₂CO₃
When lithium carbonate is heated, it decomposes as follows:
\[ \text{Li}_2\text{CO}_3 \rightarrow \text{Li}_2\text{O} + \text{CO}_2 \]
This means that for every mole of Li₂CO₃ that decomposes, 1 mole of CO₂ is produced.

### Step 4: Mass Loss Calculation
The mass loss during the reaction is due to the release of CO₂ gas. The mass loss can be expressed as:
\[ \text{Mass loss} = \text{moles of } CO_2 \times \text{molar mass of } CO_2 = X \times 44 \text{ g} \]

### Step 5: Total Initial Mass of the Mixture
The total initial mass of the mixture before heating is:
\[ \text{Total mass} = (X \times 74) + (Y \times 106) \]

### Step 6: Percentage Loss in Mass
According to the problem, the loss in mass is given as \( \frac{220}{9} \% \). Therefore, we can write the equation:
\[ \frac{X \times 44}{(X \times 74) + (Y \times 106)} = \frac{220}{9 \times 100} \]

### Step 7: Simplifying the Equation
Cross-multiplying gives:
\[ 9 \times 44X = 220 \times ((X \times 74) + (Y \times 106)) \]
\[ 396X = 220(74X + 106Y) \]
\[ 396X = 16280X + 23320Y \]
Rearranging gives:
\[ 16280X + 23320Y - 396X = 0 \]
\[ (16280 - 396)X + 23320Y = 0 \]
\[ 15884X + 23320Y = 0 \]

### Step 8: Establishing the Relationship
From the equation:
\[ 15884X = -23320Y \]
This implies:
\[ \frac{X}{Y} = -\frac{23320}{15884} \]
However, since both X and Y represent moles, we can express this in terms of a ratio:
\[ X + Y = k \quad \text{(where k is a constant)} \]
From the earlier steps, we can conclude that:
\[ X = Y \]

### Step 9: Final Molar Ratio
Thus, the molar ratio of Li₂CO₃ to Na₂CO₃ is:
\[ \frac{X}{Y} = 1:1 \]

### Conclusion
The molar ratio of \( \text{Li}_2\text{CO}_3 \) to \( \text{Na}_2\text{CO}_3 \) in the initial mixture is **1:1**.

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