Answer :
The, 2.618 grams of CO₂ will be produced when 5.00 grams of sodium bicarbonate reacts with 5.00 grams of citric acid.
To determine the amount of CO₂ gas produced when 5.00 grams of sodium bicarbonate (NaHCO₃) reacts with 5.00 grams of citric acid (C₆H₈O₇), we need to follow these steps:
1. Write the balanced chemical equation for the reaction.
2. Determine the limiting reactant by calculating the moles of each reactant.
3. Use stoichiometry to find the moles of CO₂ produced and then convert this to grams.
Step 1: Balanced Chemical Equation
The reaction between sodium bicarbonate and citric acid can be represented as:
[tex]\[ 3 \text{NaHCO}_3 + \text{C}_6\text{H}_8\text{O}_7 \rightarrow \text{Na}_3\text{C}_6\text{H}_5\text{O}_7 + 3 \text{CO}_2 + 3 \text{H}_2\text{O} \][/tex]
Step 2: Determine Moles of Each Reactant
Calculate the molar masses:
NaHCO₃: [tex]\( 23 + 1 + 12 + 3 \times 16 = 84 \)[/tex] g/mol
C₆H₈O₇: [tex]\( 6 \times 12 + 8 \times 1 + 7 \times 16 = 192 \)[/tex] g/mol
Now, calculate the moles of each reactant:
Moles of NaHCO₃: [tex]\( \frac{5.00 \, \text{g}}{84 \, \text{g/mol}} = 0.0595 \, \text{mol} \)[/tex]
Moles of C₆H₈O₇: [tex]\( \frac{5.00 \, \text{g}}{192 \, \text{g/mol}} = 0.0260 \, \text{mol} \)[/tex]
Step 3: Determine the Limiting Reactant
From the balanced equation, 3 moles of NaHCO₃ react with 1 mole of C₆H₈O₇. Therefore, we need to compare the mole ratio:
Required moles of NaHCO₃ for 0.0260 moles of C₆H₈O₇:
[tex]\[ 0.0260 \, \text{mol} \times 3 = 0.0780 \, \text{mol} \][/tex]
We only have 0.0595 moles of NaHCO₃, which is less than 0.0780 moles, so NaHCO₃ is the limiting reactant.
Step 4: Calculate Moles of CO₂ Produced
From the balanced equation, 3 moles of NaHCO₃ produce 3 moles of CO₂. Therefore, the moles of CO₂ produced are equal to the moles of NaHCO₃:
[tex]\[ {Moles ~of ~CO_2} = 0.0595 \, \text{mol} \][/tex]
Step 5: Convert Moles of CO₂ to Grams
The molar mass of CO₂ is [tex]12 + 2 \times 16 = 44[/tex] g/mol.
[tex]\[ Mass ~of~ CO_2} = 0.0595 \, \text{mol} \times 44 \, \text{g/mol} = 2.618 \, \text{g} \][/tex]
Combining 5.00 grams of sodium bicarbonate with 5.00 grams of citric acid will produce 3.43 grams of CO2 gas. This is based on the stoichiometry of the balanced chemical equation.
The reaction between sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7) produces carbon dioxide (CO2), water (H2O), and sodium citrate (Na3C6H5O7). The balanced chemical equation for this reaction is:
3 NaHCO3 + H3C6H5O7 → 3 CO2 + 3 H2O + Na3C6H5O7
Molar masses: NaHCO3 = 84 g/mol and H3C6H5O7 = 192 g/mol.
First, we convert the mass of reactants to moles:
Moles of NaHCO3 = 5.00 g / 84 g/mol = 0.0595 mol.
Moles of H3C6H5O7 = 5.00 g / 192 g/mol = 0.0260 mol.
Citric acid is the limiting reactant. According to the stoichiometry, 1 mole of H3C6H5O7 reacts with 3 moles of NaHCO3 to produce 3 moles of CO2.
Therefore, moles of CO2 = 0.0260 mol of H3C6H5O7 * 3 = 0.0780 mol.
Finally, we convert moles of CO2 to grams:
Mass of CO2 = 0.0780 mol * 44 g/mol (molar mass of CO2) = 3.43 grams.
So, 3.43 grams of CO2 gas will be produced.
