Balanced equation: NaHCO3 + HCl -> NaCl + H2O + CO2 Find the moles from 1.03 grams of CO2 produced.
The Correct Answer and Explanation is:
To find the number of moles of carbon dioxide (CO₂) produced from 1.03 grams, we use the molar mass of CO₂ and the formula:Moles=Mass (g)Molar Mass (g/mol)\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}Moles=Molar Mass (g/mol)Mass (g)
Step 1: Determine the molar mass of CO₂
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol
- CO₂ has two oxygen atoms
So,Molar Mass of CO₂=12.01+(2×16.00)=12.01+32.00=44.01 g/mol\text{Molar Mass of CO₂} = 12.01 + (2 \times 16.00) = 12.01 + 32.00 = 44.01 \, \text{g/mol}Molar Mass of CO₂=12.01+(2×16.00)=12.01+32.00=44.01g/mol
Step 2: Use the formula to find moles
Moles of CO₂=1.03 g44.01 g/mol≈0.0234 mol\text{Moles of CO₂} = \frac{1.03 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.0234 \, \text{mol}Moles of CO₂=44.01g/mol1.03g≈0.0234mol
Final Answer:
0.0234 moles of CO₂
Explanation:
The reaction between sodium bicarbonate (NaHCO₃) and hydrochloric acid (HCl) results in the formation of three products: sodium chloride (NaCl), water (H₂O), and carbon dioxide (CO₂). The balanced chemical equation is:NaHCO₃+HCl→NaCl+H₂O+CO₂\text{NaHCO₃} + \text{HCl} \rightarrow \text{NaCl} + \text{H₂O} + \text{CO₂}NaHCO₃+HCl→NaCl+H₂O+CO₂
This equation tells us that one mole of sodium bicarbonate reacts with one mole of hydrochloric acid to produce one mole each of sodium chloride, water, and carbon dioxide. Therefore, if we know the amount of CO₂ produced, we can determine the amount of NaHCO₃ that reacted.
In this problem, we are given the mass of carbon dioxide: 1.03 grams. To find the number of moles of CO₂, we must divide the mass by its molar mass. The molar mass of CO₂ is obtained by summing the atomic masses of its elements. Carbon has an atomic mass of 12.01 grams per mole, and oxygen has an atomic mass of 16.00 grams per mole. Since there are two oxygen atoms in CO₂, the molar mass becomes:12.01+2×16.00=44.01 g/mol12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol}12.01+2×16.00=44.01g/mol
Now, using the formula for moles (mass divided by molar mass), we calculate:1.0344.01≈0.0234 mol\frac{1.03}{44.01} \approx 0.0234 \, \text{mol}44.011.03≈0.0234mol
Thus, 1.03 grams of carbon dioxide is equivalent to 0.0234 moles, and this value reflects the amount of CO₂ produced in the reaction.
