Certainly! Let's break down the problem step by step.
1. Identify Important Information:
- Volume of HCl solution [tex]\(\left( \mathrm{HCl(aq)} \right)\)[/tex]: 50.0 mL
- Molarity of HCl solution: 2.0 M (Moles per liter)
2. Convert the Volume from Milliliters to Liters:
[tex]\[
\text{Volume in Liters} = \frac{\text{Volume in Milliliters}}{1000}
\][/tex]
Substituting the given volume:
[tex]\[
\text{Volume in Liters} = \frac{50.0 \text{ mL}}{1000} = 0.050 \text{ L}
\][/tex]
3. Calculate Moles of HCl:
Molarity (M) is defined as the number of moles of solute per liter of solution. Therefore, we can use the formula:
[tex]\[
\text{Moles of HCl} = \text{Molarity} \times \text{Volume (in Liters)}
\][/tex]
Substituting the values:
[tex]\[
\text{Moles of HCl} = 2.0 \text{ M} \times 0.050 \text{ L} = 0.10 \text{ moles}
\][/tex]
4. Determine the Moles of [tex]\( CO_2 \)[/tex] Produced:
According to the balanced chemical equation:
[tex]\[
2 \, \mathrm{HCl(aq)} + \mathrm{Na_2CO_3(aq)} \rightarrow 2 \, \mathrm{NaCl(aq)} + \mathrm{H_2O(l)} + \mathrm{CO_2(g)}
\][/tex]
- 2 moles of HCl produce 1 mole of [tex]\( CO_2 \)[/tex].
Therefore, the moles of [tex]\( CO_2 \)[/tex] produced can be found by dividing the moles of HCl by 2:
[tex]\[
\text{Moles of \( CO_2 \)} = \frac{\text{Moles of HCl}}{2} = \frac{0.10 \text{ moles}}{2} = 0.05 \text{ moles}
\][/tex]
So, when 50.0 mL of 2.0 M HCl reacts with excess Na₂CO₃, 0.05 moles of [tex]\( CO_2 \)[/tex] are produced.
