200cm$^3$ of 0.50mol/dm$^3$ solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is

1. Analyze the Question

We are given the volume ($200\text{ cm}^3$) and concentration ($0.50\text{ mol/dm}^3$) of a solution of calcium hydrogen trioxocarbonate(IV), $\text{Ca(HCO}_3)_2$. We need to find the mass of the solid precipitate formed when this solution is heated.

2. Explain the Concept

When calcium hydrogen trioxocarbonate(IV) is heated, it undergoes thermal decomposition to form calcium trioxocarbonate(IV) (which is the solid precipitate), water, and carbon(IV) oxide gas.

The balanced chemical equation is:

From the equation, 1 mole of $\text{Ca(HCO}_3)_2$ produces 1 mole of $\text{CaCO}_3$.

3. Show the Solution

Step 1: Calculate the number of moles of $\text{Ca(HCO}_3)_2$

• Volume ($V$) = $200\text{ cm}^3 = 0.2\text{ dm}^3$
• Concentration ($C$) = $0.50\text{ mol/dm}^3$
• $\text{Moles} = C \times V = 0.50 \times 0.2 = 0.1\text{ moles}$

Step 2: Determine the moles of $\text{CaCO}_3$ produced Since the mole ratio of $\text{Ca(HCO}_3)_2$ to $\text{CaCO}_3$ is $1:1$, $0.1\text{ moles}$ of $\text{Ca(HCO}_3)_2$ will produce $0.1\text{ moles}$ of $\text{CaCO}_3$.

Step 3: Calculate the molar mass of $\text{CaCO}_3$

• $\text{Ca} = 40, \text{C} = 12, \text{O} = 16$
• $\text{Molar mass} = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100\text{ g/mol}$

Step 4: Calculate the mass of the precipitate

• $\text{Mass} = \text{moles} \times \text{molar mass}$
• $\text{Mass} = 0.1\text{ mol} \times 100\text{ g/mol} = 10\text{ g}$

4. Identify the Answer

The maximum weight of the solid precipitated is $10\text{ g}$. This matches option A.

5. Tip

Always remember to convert your volume from $\text{cm}^3$ to $\text{dm}^3$ (by dividing by 1000) before calculating moles. Also, ensure your chemical equation is balanced before using the mole ratio to solve for the product!