Answer :
Step-by-Step Solution:
1. Balancing the Chemical Equation:
The given chemical equation is:
[tex]\[ \text{CaCl}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{NaCl} + \text{CaCO}_3 \][/tex]
To balance this equation, we need to make sure the number of atoms for each element on the reactant side is equal to the number of atoms on the product side.
- There is 1 calcium (Ca) atom on both sides.
- There are 2 chlorine (Cl) atoms on the reactant side but only 1 chlorine atom on the product side.
- There are 2 sodium (Na) atoms on the reactant side but only 1 sodium atom on the product side.
- There is 1 carbonate (CO₃) group on both sides.
To balance the chlorine and sodium atoms, we place a coefficient of 2 in front of NaCl:
[tex]\[ \text{CaCl}_2 + \text{Na}_2\text{CO}_3 \rightarrow 2 \text{NaCl} + \text{CaCO}_3 \][/tex]
Now the equation is balanced.
2. Moles of Sodium Carbonate:
We are given a 29.35 gram sample of sodium carbonate ([tex]\(\text{Na}_2\text{CO}_3\)[/tex]).
The molar mass of sodium carbonate ([tex]\(\text{Na}_2\text{CO}_3\)[/tex]) is 105.99 g/mol.
The number of moles of sodium carbonate can be calculated using the formula:
[tex]\[ \text{Moles of } \text{Na}_2\text{CO}_3 = \frac{\text{mass of } \text{Na}_2\text{CO}_3}{\text{molar mass of } \text{Na}_2\text{CO}_3} \][/tex]
Substituting the given values:
[tex]\[ \text{Moles of } \text{Na}_2\text{CO}_3 = \frac{29.35 \text{ g}}{105.99 \text{ g/mol}} \approx 0.2769 \text{ moles} \][/tex]
3. Moles of Sodium Chloride:
According to the balanced chemical equation, 1 mole of sodium carbonate ([tex]\(\text{Na}_2\text{CO}_3\)[/tex]) produces 2 moles of sodium chloride ([tex]\(\text{NaCl}\)[/tex]).
Therefore, the moles of sodium chloride produced from the reaction will be:
[tex]\[ \text{Moles of } \text{NaCl} = \text{Moles of } \text{Na}_2\text{CO}_3 \times 2 \][/tex]
Substituting the calculated moles of sodium carbonate:
[tex]\[ \text{Moles of } \text{NaCl} = 0.2769 \text{ moles} \times 2 \approx 0.5538 \text{ moles} \][/tex]
4. Mass of Sodium Chloride:
The molar mass of sodium chloride ([tex]\(\text{NaCl}\)[/tex]) is 58.44 g/mol.
The mass of sodium chloride produced can be calculated using the formula:
[tex]\[ \text{Mass of } \text{NaCl} = \text{Moles of } \text{NaCl} \times \text{Molar mass of } \text{NaCl} \][/tex]
Substituting the calculated moles and molar mass:
[tex]\[ \text{Mass of } \text{NaCl} = 0.5538 \text{ moles} \times 58.44 \text{ g/mol} \approx 32.37 \text{ grams} \][/tex]
Final Answer:
[tex]\[ 32.37 \text{ grams} \][/tex]
This is how we determine the amount of sodium chloride produced from a 29.35 gram sample of sodium carbonate in this reaction.
1. Balancing the Chemical Equation:
The given chemical equation is:
[tex]\[ \text{CaCl}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{NaCl} + \text{CaCO}_3 \][/tex]
To balance this equation, we need to make sure the number of atoms for each element on the reactant side is equal to the number of atoms on the product side.
- There is 1 calcium (Ca) atom on both sides.
- There are 2 chlorine (Cl) atoms on the reactant side but only 1 chlorine atom on the product side.
- There are 2 sodium (Na) atoms on the reactant side but only 1 sodium atom on the product side.
- There is 1 carbonate (CO₃) group on both sides.
To balance the chlorine and sodium atoms, we place a coefficient of 2 in front of NaCl:
[tex]\[ \text{CaCl}_2 + \text{Na}_2\text{CO}_3 \rightarrow 2 \text{NaCl} + \text{CaCO}_3 \][/tex]
Now the equation is balanced.
2. Moles of Sodium Carbonate:
We are given a 29.35 gram sample of sodium carbonate ([tex]\(\text{Na}_2\text{CO}_3\)[/tex]).
The molar mass of sodium carbonate ([tex]\(\text{Na}_2\text{CO}_3\)[/tex]) is 105.99 g/mol.
The number of moles of sodium carbonate can be calculated using the formula:
[tex]\[ \text{Moles of } \text{Na}_2\text{CO}_3 = \frac{\text{mass of } \text{Na}_2\text{CO}_3}{\text{molar mass of } \text{Na}_2\text{CO}_3} \][/tex]
Substituting the given values:
[tex]\[ \text{Moles of } \text{Na}_2\text{CO}_3 = \frac{29.35 \text{ g}}{105.99 \text{ g/mol}} \approx 0.2769 \text{ moles} \][/tex]
3. Moles of Sodium Chloride:
According to the balanced chemical equation, 1 mole of sodium carbonate ([tex]\(\text{Na}_2\text{CO}_3\)[/tex]) produces 2 moles of sodium chloride ([tex]\(\text{NaCl}\)[/tex]).
Therefore, the moles of sodium chloride produced from the reaction will be:
[tex]\[ \text{Moles of } \text{NaCl} = \text{Moles of } \text{Na}_2\text{CO}_3 \times 2 \][/tex]
Substituting the calculated moles of sodium carbonate:
[tex]\[ \text{Moles of } \text{NaCl} = 0.2769 \text{ moles} \times 2 \approx 0.5538 \text{ moles} \][/tex]
4. Mass of Sodium Chloride:
The molar mass of sodium chloride ([tex]\(\text{NaCl}\)[/tex]) is 58.44 g/mol.
The mass of sodium chloride produced can be calculated using the formula:
[tex]\[ \text{Mass of } \text{NaCl} = \text{Moles of } \text{NaCl} \times \text{Molar mass of } \text{NaCl} \][/tex]
Substituting the calculated moles and molar mass:
[tex]\[ \text{Mass of } \text{NaCl} = 0.5538 \text{ moles} \times 58.44 \text{ g/mol} \approx 32.37 \text{ grams} \][/tex]
Final Answer:
[tex]\[ 32.37 \text{ grams} \][/tex]
This is how we determine the amount of sodium chloride produced from a 29.35 gram sample of sodium carbonate in this reaction.
