Answer :
Let's solve the given question step-by-step:
### Part A: Moles of NaCl Produced
The balanced chemical equation for the reaction is:
[tex]\[ NaHCO_3(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l) + CO_2(g) \][/tex]
From the balanced equation, you can see that the molar ratio of [tex]\( NaHCO_3 \)[/tex] to [tex]\( CO_2 \)[/tex] and [tex]\( NaCl \)[/tex] to [tex]\( CO_2 \)[/tex] is 1:1. This means 1 mole of [tex]\( NaHCO_3 \)[/tex] produces 1 mole of [tex]\( CO_2 \)[/tex] and 1 mole of [tex]\( NaCl \)[/tex].
Given that 4.5 × 10^{-2} moles of [tex]\( CO_2 \)[/tex] are produced, the moles of [tex]\( NaCl \)[/tex] produced will be the same due to the 1:1 ratio:
[tex]\[ \text{Moles of } NaCl = 4.5 \times 10^{-2} \, \text{moles} \][/tex]
Thus, 0.045 moles of NaCl are produced.
### Part B: Molar Mass of NaCl
To find the molar mass of [tex]\( NaCl \)[/tex], we need to add the atomic masses of sodium (Na) and chlorine (Cl):
- Molar mass of [tex]\( Na \)[/tex] (Sodium) = 22.99 g/mol
- Molar mass of [tex]\( Cl \)[/tex] (Chlorine) = 35.45 g/mol
Adding these, we get the molar mass of [tex]\( NaCl \)[/tex]:
[tex]\[ \text{Molar mass of } NaCl = 22.99 + 35.45 = 58.44 \, \text{g/mol} \][/tex]
Thus, the molar mass of NaCl is 58.44 g/mol.
### Part C: Molar Mass of Sodium Bicarbonate ([tex]\( NaHCO_3 \)[/tex])
To calculate the molar mass of [tex]\( NaHCO_3 \)[/tex], we add the atomic masses of sodium (Na), hydrogen (H), carbon (C), and oxygen (O):
- Molar mass of [tex]\( Na \)[/tex] (Sodium) = 22.99 g/mol
- Molar mass of [tex]\( H \)[/tex] (Hydrogen) = 1.01 g/mol
- Molar mass of [tex]\( C \)[/tex] (Carbon) = 12.01 g/mol
- Molar mass of [tex]\( O \)[/tex] (Oxygen) = 16.00 g/mol (and since there are three oxygen atoms, we multiply by 3)
So the calculation for the molar mass of [tex]\( NaHCO_3 \)[/tex] is:
[tex]\[ \text{Molar mass of } NaHCO_3 = 22.99 + 1.01 + 12.01 + (3 \times 16.00) = 22.99 + 1.01 + 12.01 + 48.00 = 84.01 \, \text{g/mol} \][/tex]
Thus, the molar mass of NaHCO_3 is 84.01 g/mol.
### Part D: Grams of Sodium Chloride Formed
We are given 80.0 grams of sodium bicarbonate ([tex]\( NaHCO_3 \)[/tex]). To find out how many grams of sodium chloride ([tex]\( NaCl \)[/tex]) are formed, we need to follow these steps:
1. Calculate the moles of [tex]\( NaHCO_3 \)[/tex] present in 80.0 grams:
[tex]\[ \text{Moles of } NaHCO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{80.0 \, \text{g}}{84.01 \, \text{g/mol}} = 0.952 \, \text{moles} \][/tex]
This step ensures we know how many moles of [tex]\( NaHCO_3 \)[/tex] we are starting with.
2. Use the stoichiometry of the balanced equation to determine the moles of [tex]\( NaCl \)[/tex] formed:
From the balanced equation and the 1:1 ratio, 1 mole of [tex]\( NaHCO_3 \)[/tex] produces 1 mole of [tex]\( NaCl \)[/tex]. Therefore, 0.952 moles of [tex]\( NaHCO_3 \)[/tex] will produce 0.952 moles of [tex]\( NaCl \)[/tex].
3. Calculate the mass of [tex]\( NaCl \)[/tex] formed from the moles of [tex]\( NaCl \)[/tex]:
[tex]\[ \text{Mass of } NaCl = \text{moles} \times \text{molar mass} = 0.952 \, \text{moles} \times 58.44 \, \text{g/mol} \approx 55.65 \, \text{grams} \][/tex]
Thus, 55.65 grams of NaCl are formed from the decomposition of 80.0 grams of sodium bicarbonate.
### Part A: Moles of NaCl Produced
The balanced chemical equation for the reaction is:
[tex]\[ NaHCO_3(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l) + CO_2(g) \][/tex]
From the balanced equation, you can see that the molar ratio of [tex]\( NaHCO_3 \)[/tex] to [tex]\( CO_2 \)[/tex] and [tex]\( NaCl \)[/tex] to [tex]\( CO_2 \)[/tex] is 1:1. This means 1 mole of [tex]\( NaHCO_3 \)[/tex] produces 1 mole of [tex]\( CO_2 \)[/tex] and 1 mole of [tex]\( NaCl \)[/tex].
Given that 4.5 × 10^{-2} moles of [tex]\( CO_2 \)[/tex] are produced, the moles of [tex]\( NaCl \)[/tex] produced will be the same due to the 1:1 ratio:
[tex]\[ \text{Moles of } NaCl = 4.5 \times 10^{-2} \, \text{moles} \][/tex]
Thus, 0.045 moles of NaCl are produced.
### Part B: Molar Mass of NaCl
To find the molar mass of [tex]\( NaCl \)[/tex], we need to add the atomic masses of sodium (Na) and chlorine (Cl):
- Molar mass of [tex]\( Na \)[/tex] (Sodium) = 22.99 g/mol
- Molar mass of [tex]\( Cl \)[/tex] (Chlorine) = 35.45 g/mol
Adding these, we get the molar mass of [tex]\( NaCl \)[/tex]:
[tex]\[ \text{Molar mass of } NaCl = 22.99 + 35.45 = 58.44 \, \text{g/mol} \][/tex]
Thus, the molar mass of NaCl is 58.44 g/mol.
### Part C: Molar Mass of Sodium Bicarbonate ([tex]\( NaHCO_3 \)[/tex])
To calculate the molar mass of [tex]\( NaHCO_3 \)[/tex], we add the atomic masses of sodium (Na), hydrogen (H), carbon (C), and oxygen (O):
- Molar mass of [tex]\( Na \)[/tex] (Sodium) = 22.99 g/mol
- Molar mass of [tex]\( H \)[/tex] (Hydrogen) = 1.01 g/mol
- Molar mass of [tex]\( C \)[/tex] (Carbon) = 12.01 g/mol
- Molar mass of [tex]\( O \)[/tex] (Oxygen) = 16.00 g/mol (and since there are three oxygen atoms, we multiply by 3)
So the calculation for the molar mass of [tex]\( NaHCO_3 \)[/tex] is:
[tex]\[ \text{Molar mass of } NaHCO_3 = 22.99 + 1.01 + 12.01 + (3 \times 16.00) = 22.99 + 1.01 + 12.01 + 48.00 = 84.01 \, \text{g/mol} \][/tex]
Thus, the molar mass of NaHCO_3 is 84.01 g/mol.
### Part D: Grams of Sodium Chloride Formed
We are given 80.0 grams of sodium bicarbonate ([tex]\( NaHCO_3 \)[/tex]). To find out how many grams of sodium chloride ([tex]\( NaCl \)[/tex]) are formed, we need to follow these steps:
1. Calculate the moles of [tex]\( NaHCO_3 \)[/tex] present in 80.0 grams:
[tex]\[ \text{Moles of } NaHCO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{80.0 \, \text{g}}{84.01 \, \text{g/mol}} = 0.952 \, \text{moles} \][/tex]
This step ensures we know how many moles of [tex]\( NaHCO_3 \)[/tex] we are starting with.
2. Use the stoichiometry of the balanced equation to determine the moles of [tex]\( NaCl \)[/tex] formed:
From the balanced equation and the 1:1 ratio, 1 mole of [tex]\( NaHCO_3 \)[/tex] produces 1 mole of [tex]\( NaCl \)[/tex]. Therefore, 0.952 moles of [tex]\( NaHCO_3 \)[/tex] will produce 0.952 moles of [tex]\( NaCl \)[/tex].
3. Calculate the mass of [tex]\( NaCl \)[/tex] formed from the moles of [tex]\( NaCl \)[/tex]:
[tex]\[ \text{Mass of } NaCl = \text{moles} \times \text{molar mass} = 0.952 \, \text{moles} \times 58.44 \, \text{g/mol} \approx 55.65 \, \text{grams} \][/tex]
Thus, 55.65 grams of NaCl are formed from the decomposition of 80.0 grams of sodium bicarbonate.
