Final answer:
To find the volume of 0.600 M HCl needed to react with 2.50 g of NaHCO3, we first calculate the number of moles of NaHCO3 and then use the molarity of HCl to find the volume. The calculation yields approximately 50.00 mL of HCl, corresponding to option d.
Step-by-step explanation:
The student's question asks, "What volume of 0.600 M HCl is required to react completely with 2.50 g of sodium hydrogen carbonate (NaHCO3)?" To answer this, we will need to calculate the number of moles of NaHCO3 and use the stoichiometry of the balanced chemical reaction to find the volume of HCl needed.
First, we calculate the molar mass of NaHCO3:
- Na: 1 x 22.99 g/mol
- H: 1 x 1.01 g/mol
- C: 1 x 12.01 g/mol
- O: 3 x 16.00 g/mol
Molar mass of NaHCO3 = 22.99 + 1.01 + 12.01 + (3 x 16.00) = 84.01 g/mol
Next, we calculate the number of moles of NaHCO3 we have:
2.50 g NaHCO3 x (1 mol NaHCO3 / 84.01 g NaHCO3) = 0.0298 mol NaHCO3
According to the balanced chemical equation, 1 mol of NaHCO3 reacts with 1 mol of HCl. Therefore, we need 0.0298 mol of HCl. Now we use the molarity of HCl to find the volume needed:
0.0298 mol HCl x (1 L HCl / 0.600 mol HCl) = 0.0497 L of HCl
Converting liters to milliliters:
0.0497 L x (1000 mL / 1 L) = 49.7 mL
Therefore, the correct volume of 0.600 M HCl required to react completely with 2.50 g of NaHCO3 is approximately 50.00 mL, which is closest to choice d) 50.00 mL.