Questions: Na2CO3(aq) + CaCl2(aq) → 2 NaCl(aq) + CaCO3(s) Calculate the volume (in mL) of 0.100 M Na2CO3 needed to produce 1.00 g of CaCO3(s) There is an excess of CaCl2. Molar mass of calcium carbonate = 100.09 g / mol Volume of sodium carbonate = mL

Na2CO3(aq) + CaCl2(aq) → 2 NaCl(aq) + CaCO3(s)

Calculate the volume (in mL) of 0.100 M Na2CO3 needed to produce 1.00 g of CaCO3(s)

There is an excess of CaCl2.
Molar mass of calcium carbonate = 100.09 g / mol

Volume of sodium carbonate = mL

Transcript text: 2) \[ \mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CaCl}_{2}(a q) \rightarrow 2 \mathrm{NaCl}(a q)+\mathrm{CaCO}_{3}(s) \] Calculate the volume (in mL ) of $0.100 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}$ needed to produce $\mathbf{1 . 0 0} \mathrm{g}$ of \[ \mathrm{CaCO}_{3}(s) \] There is an excess of $\mathrm{CaCl}_{2}$. Molar mass of calcium carbonate $=100.09 \mathrm{~g} / \mathrm{mol}$ Volume of sodium carbonate $=$ $\square$ mL

Solution

Solution Steps

Step 1: Determine the Moles of Calcium Carbonate Required

First, calculate the number of moles of calcium carbonate ($\text{CaCO}_3$) needed to produce 1.00 g of $\text{CaCO}_3$.

\[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{1.00 \, \text{g}}{100.09 \, \text{g/mol}} = 0.009991 \, \text{mol} \]

Step 2: Use Stoichiometry to Find Moles of Sodium Carbonate Needed

From the balanced chemical equation:

\[ \text{Na}_2\text{CO}_3(aq) + \text{CaCl}_2(aq) \rightarrow 2\text{NaCl}(aq) + \text{CaCO}_3(s) \]

The stoichiometry shows that 1 mole of $\text{Na}_2\text{CO}_3$ produces 1 mole of $\text{CaCO}_3$. Therefore, the moles of $\text{Na}_2\text{CO}_3$ needed are the same as the moles of $\text{CaCO}_3$.

\[ \text{Moles of Na}_2\text{CO}_3 = 0.009991 \, \text{mol} \]

Step 3: Calculate the Volume of Sodium Carbonate Solution Required

Given the concentration of the $\text{Na}_2\text{CO}_3$ solution is 0.100 M, use the formula for molarity to find the volume:

\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

Rearranging to find the volume:

\[ \text{Volume in liters} = \frac{\text{moles of Na}_2\text{CO}_3}{\text{Molarity}} = \frac{0.009991 \, \text{mol}}{0.100 \, \text{mol/L}} = 0.09991 \, \text{L} \]

Convert the volume from liters to milliliters:

\[ \text{Volume in mL} = 0.09991 \, \text{L} \times 1000 \, \text{mL/L} = 99.91 \, \text{mL} \]