Certainly! Let's solve this question step-by-step.
1. Identify the reaction and the data given:
- We are given the chemical reaction:
[tex]\[
NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2
\][/tex]
- We know that the container holds 0.0135 moles of [tex]\( CO_2 \)[/tex].
- The molar mass of [tex]\( NaHCO_3 \)[/tex] is given as 84.01 g/mol.
2. Stoichiometric relationship:
- According to the balanced chemical equation, 1 mole of [tex]\( NaHCO_3 \)[/tex] produces 1 mole of [tex]\( CO_2 \)[/tex].
- Thus, the number of moles of [tex]\( NaHCO_3 \)[/tex] needed to produce 0.0135 moles of [tex]\( CO_2 \)[/tex] is also 0.0135 moles.
3. Calculate the mass of [tex]\( NaHCO_3 \)[/tex]:
- The molar mass of [tex]\( NaHCO_3 \)[/tex] is 84.01 g/mol. This means 1 mole of [tex]\( NaHCO_3 \)[/tex] weighs 84.01 grams.
- To find the mass of [tex]\( NaHCO_3 \)[/tex] needed, we use the number of moles and the molar mass:
[tex]\[
\text{mass of } NaHCO_3 = \text{moles of } NaHCO_3 \times \text{molar mass of } NaHCO_3
\][/tex]
Substituting the values:
[tex]\[
\text{mass of } NaHCO_3 = 0.0135 \text{ moles} \times 84.01 \text{ g/mol}
\][/tex]
4. Perform the multiplication:
- Calculating the above expression:
[tex]\[
\text{mass of } NaHCO_3 = 0.0135 \times 84.01 = 1.134135 \text{ g}
\][/tex]
So, the mass of [tex]\( NaHCO_3 \)[/tex] needed to generate 0.0135 moles of [tex]\( CO_2 \)[/tex] is approximately 1.134 grams.
