Answer :
Let's solve the given problem step-by-step:
### (a) Find the limiting reactant
The balanced chemical equation for the reaction between [tex]\( \text{CaCO}_3 \)[/tex] and [tex]\( \text{HCl} \)[/tex] is:
[tex]\[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \][/tex]
Step 1: Calculate the number of moles of [tex]\( \text{CaCO}_3 \)[/tex] and [tex]\( \text{HCl} \)[/tex].
Molar mass of [tex]\( \text{CaCO}_3 \)[/tex]:
[tex]\[ 40 \, (\text{Ca}) + 12 \, (\text{C}) + 3 \times 16 \, (\text{O}) = 100 \, \text{g/mol} \][/tex]
Number of moles of [tex]\( \text{CaCO}_3 \)[/tex]:
[tex]\[ \frac{10 \, \text{g}}{100 \, \text{g/mol}} = 0.1 \, \text{moles} \][/tex]
Molar mass of [tex]\( \text{HCl} \)[/tex]:
[tex]\[ 1 \, (\text{H}) + 35.5 \, (\text{Cl}) = 36.5 \, \text{g/mol} \][/tex]
Number of moles of [tex]\( \text{HCl} \)[/tex]:
[tex]\[ \frac{7.665 \, \text{g}}{36.5 \, \text{g/mol}} \approx 0.21 \, \text{moles} \][/tex]
Step 2: Determine the limiting reactant by comparing the mole ratio from the balanced equation.
From the balanced equation, 1 mole of [tex]\( \text{CaCO}_3 \)[/tex] reacts with 2 moles of [tex]\( \text{HCl} \)[/tex].
So, for 0.1 moles of [tex]\( \text{CaCO}_3 \)[/tex], we need [tex]\( 0.1 \times 2 = 0.2 \, \text{moles} \)[/tex] of [tex]\( \text{HCl} \)[/tex].
We have 0.21 moles of [tex]\( \text{HCl} \)[/tex], which is more than 0.2 moles, meaning [tex]\( \text{CaCO}_3 \)[/tex] is the limiting reactant.
[tex]\[ \text{Limiting reactant:} \ \text{CaCO}_3 \][/tex]
### (b) Calculate the number of moles of reactant left over unreacted.
Since [tex]\( \text{CaCO}_3 \)[/tex] is the limiting reactant, it will be completely consumed, and some [tex]\( \text{HCl} \)[/tex] will be left unreacted.
Moles of [tex]\( \text{HCl} \)[/tex] used:
[tex]\[ 2 \times \text{moles of } \text{CaCO}_3 = 2 \times 0.1 = 0.2 \, \text{moles} \][/tex]
Moles of [tex]\( \text{HCl} \)[/tex] left unreacted:
[tex]\[ 0.21 - 0.2 = 0.01 \, \text{moles} \][/tex]
### (c) Calculate the volume of [tex]\( \text{CO}_2 \)[/tex] gas produced at NTP.
From the balanced equation, 1 mole of [tex]\( \text{CaCO}_3 \)[/tex] produces 1 mole of [tex]\( \text{CO}_2 \)[/tex].
Moles of [tex]\( \text{CO}_2 \)[/tex] produced:
[tex]\[ \text{moles of } \text{CaCO}_3 = 0.1 \, \text{moles} \][/tex]
At NTP (Normal Temperature and Pressure), 1 mole of gas occupies 22.4 liters.
Volume of [tex]\( \text{CO}_2 \)[/tex] produced:
[tex]\[ 0.1 \times 22.4 \, \text{liters} = 2.24 \, \text{liters} \][/tex]
### (d) Calculate the number of grams of NaOH required to absorb the [tex]\( \text{CO}_2 \)[/tex] gas produced as [tex]\( \text{Na}_2\text{CO}_3 \)[/tex].
The balanced reaction for the absorption of [tex]\( \text{CO}_2 \)[/tex] by [tex]\( \text{NaOH} \)[/tex] is:
[tex]\[ \text{CO}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \][/tex]
From the equation, 1 mole of [tex]\( \text{CO}_2 \)[/tex] requires 2 moles of [tex]\( \text{NaOH} \)[/tex].
Moles of [tex]\( \text{NaOH} \)[/tex] required:
[tex]\[ 2 \times \text{moles of } \text{CO}_2 = 2 \times 0.1 = 0.2 \, \text{moles} \][/tex]
Molar mass of [tex]\( \text{NaOH} \)[/tex]:
[tex]\[ 23 \, (\text{Na}) + 16 \, (\text{O}) + 1 \, (\text{H}) = 40 \, \text{g/mol} \][/tex]
Grams of [tex]\( \text{NaOH} \)[/tex] required:
[tex]\[ 0.2 \times 40 \, \text{g/mol} = 8 \, \text{grams} \][/tex]
### Summary
(a) Limiting reactant: [tex]\( \text{CaCO}_3 \)[/tex]
(b) Moles of [tex]\( \text{HCl} \)[/tex] left unreacted: [tex]\( 0.01 \)[/tex] moles
(c) Volume of [tex]\( \text{CO}_2 \)[/tex] produced at NTP: [tex]\( 2.24 \)[/tex] liters
(d) Grams of [tex]\( \text{NaOH} \)[/tex] required: [tex]\( 8 \)[/tex] grams
### (a) Find the limiting reactant
The balanced chemical equation for the reaction between [tex]\( \text{CaCO}_3 \)[/tex] and [tex]\( \text{HCl} \)[/tex] is:
[tex]\[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \][/tex]
Step 1: Calculate the number of moles of [tex]\( \text{CaCO}_3 \)[/tex] and [tex]\( \text{HCl} \)[/tex].
Molar mass of [tex]\( \text{CaCO}_3 \)[/tex]:
[tex]\[ 40 \, (\text{Ca}) + 12 \, (\text{C}) + 3 \times 16 \, (\text{O}) = 100 \, \text{g/mol} \][/tex]
Number of moles of [tex]\( \text{CaCO}_3 \)[/tex]:
[tex]\[ \frac{10 \, \text{g}}{100 \, \text{g/mol}} = 0.1 \, \text{moles} \][/tex]
Molar mass of [tex]\( \text{HCl} \)[/tex]:
[tex]\[ 1 \, (\text{H}) + 35.5 \, (\text{Cl}) = 36.5 \, \text{g/mol} \][/tex]
Number of moles of [tex]\( \text{HCl} \)[/tex]:
[tex]\[ \frac{7.665 \, \text{g}}{36.5 \, \text{g/mol}} \approx 0.21 \, \text{moles} \][/tex]
Step 2: Determine the limiting reactant by comparing the mole ratio from the balanced equation.
From the balanced equation, 1 mole of [tex]\( \text{CaCO}_3 \)[/tex] reacts with 2 moles of [tex]\( \text{HCl} \)[/tex].
So, for 0.1 moles of [tex]\( \text{CaCO}_3 \)[/tex], we need [tex]\( 0.1 \times 2 = 0.2 \, \text{moles} \)[/tex] of [tex]\( \text{HCl} \)[/tex].
We have 0.21 moles of [tex]\( \text{HCl} \)[/tex], which is more than 0.2 moles, meaning [tex]\( \text{CaCO}_3 \)[/tex] is the limiting reactant.
[tex]\[ \text{Limiting reactant:} \ \text{CaCO}_3 \][/tex]
### (b) Calculate the number of moles of reactant left over unreacted.
Since [tex]\( \text{CaCO}_3 \)[/tex] is the limiting reactant, it will be completely consumed, and some [tex]\( \text{HCl} \)[/tex] will be left unreacted.
Moles of [tex]\( \text{HCl} \)[/tex] used:
[tex]\[ 2 \times \text{moles of } \text{CaCO}_3 = 2 \times 0.1 = 0.2 \, \text{moles} \][/tex]
Moles of [tex]\( \text{HCl} \)[/tex] left unreacted:
[tex]\[ 0.21 - 0.2 = 0.01 \, \text{moles} \][/tex]
### (c) Calculate the volume of [tex]\( \text{CO}_2 \)[/tex] gas produced at NTP.
From the balanced equation, 1 mole of [tex]\( \text{CaCO}_3 \)[/tex] produces 1 mole of [tex]\( \text{CO}_2 \)[/tex].
Moles of [tex]\( \text{CO}_2 \)[/tex] produced:
[tex]\[ \text{moles of } \text{CaCO}_3 = 0.1 \, \text{moles} \][/tex]
At NTP (Normal Temperature and Pressure), 1 mole of gas occupies 22.4 liters.
Volume of [tex]\( \text{CO}_2 \)[/tex] produced:
[tex]\[ 0.1 \times 22.4 \, \text{liters} = 2.24 \, \text{liters} \][/tex]
### (d) Calculate the number of grams of NaOH required to absorb the [tex]\( \text{CO}_2 \)[/tex] gas produced as [tex]\( \text{Na}_2\text{CO}_3 \)[/tex].
The balanced reaction for the absorption of [tex]\( \text{CO}_2 \)[/tex] by [tex]\( \text{NaOH} \)[/tex] is:
[tex]\[ \text{CO}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \][/tex]
From the equation, 1 mole of [tex]\( \text{CO}_2 \)[/tex] requires 2 moles of [tex]\( \text{NaOH} \)[/tex].
Moles of [tex]\( \text{NaOH} \)[/tex] required:
[tex]\[ 2 \times \text{moles of } \text{CO}_2 = 2 \times 0.1 = 0.2 \, \text{moles} \][/tex]
Molar mass of [tex]\( \text{NaOH} \)[/tex]:
[tex]\[ 23 \, (\text{Na}) + 16 \, (\text{O}) + 1 \, (\text{H}) = 40 \, \text{g/mol} \][/tex]
Grams of [tex]\( \text{NaOH} \)[/tex] required:
[tex]\[ 0.2 \times 40 \, \text{g/mol} = 8 \, \text{grams} \][/tex]
### Summary
(a) Limiting reactant: [tex]\( \text{CaCO}_3 \)[/tex]
(b) Moles of [tex]\( \text{HCl} \)[/tex] left unreacted: [tex]\( 0.01 \)[/tex] moles
(c) Volume of [tex]\( \text{CO}_2 \)[/tex] produced at NTP: [tex]\( 2.24 \)[/tex] liters
(d) Grams of [tex]\( \text{NaOH} \)[/tex] required: [tex]\( 8 \)[/tex] grams
