Questions: Calculate the volume (in mL) of 0.100 M CaCl2 needed to produce 2.00 g of CaCO3(s). There is an excess of Na2CO3. Molar mass of calcium carbonate = 100.09 g / mol

Calculate the volume (in mL) of 0.100 M CaCl2 needed to produce 2.00 g of CaCO3(s). There is an excess of Na2CO3.
Molar mass of calcium carbonate = 100.09 g / mol

Transcript text: Calculate the volume (in mL ) of $0.100 \mathrm{M} \mathrm{CaCl}_{2}$ needed to produce 2.00 g of $\mathrm{CaCO}_{3}(\mathrm{~s})$. There is an excess of $\mathrm{Na}_{2} \mathrm{CO}_{3}$. Molar mass of calcium carbonate $=100.09 \mathrm{~g} / \mathrm{mol}$

Solution

Solution Steps

Step 1: Determine Moles of $\mathrm{CaCO}_3$

First, calculate the number of moles of $\mathrm{CaCO}_3$ using its given mass and molar mass.

\[ \text{Moles of } \mathrm{CaCO}_3 = \frac{\text{Mass of } \mathrm{CaCO}_3}{\text{Molar mass of } \mathrm{CaCO}_3} = \frac{2.00 \, \text{g}}{100.09 \, \text{g/mol}} \]

\[ \text{Moles of } \mathrm{CaCO}_3 = 0.01998 \, \text{mol} \]

Step 2: Use Stoichiometry to Find Moles of $\mathrm{CaCl}_2$

The balanced chemical equation for the reaction is:

\[ \mathrm{CaCl}_2 + \mathrm{Na}_2\mathrm{CO}_3 \rightarrow \mathrm{CaCO}_3 + 2\mathrm{NaCl} \]

From the equation, 1 mole of $\mathrm{CaCl}_2$ produces 1 mole of $\mathrm{CaCO}_3$. Therefore, the moles of $\mathrm{CaCl}_2$ needed are equal to the moles of $\mathrm{CaCO}_3$.

\[ \text{Moles of } \mathrm{CaCl}_2 = 0.01998 \, \text{mol} \]

Step 3: Calculate Volume of $\mathrm{CaCl}_2$ Solution

Use the molarity of the $\mathrm{CaCl}_2$ solution to find the volume needed.

\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]

Rearranging for volume:

\[ \text{Volume of solution in liters} = \frac{\text{Moles of } \mathrm{CaCl}_2}{\text{Molarity}} = \frac{0.01998 \, \text{mol}}{0.100 \, \text{M}} \]

\[ \text{Volume of solution in liters} = 0.1998 \, \text{L} \]

Convert liters to milliliters:

\[ \text{Volume in mL} = 0.1998 \, \text{L} \times 1000 \, \text{mL/L} = 199.8 \, \text{mL} \]