what is a volume 2 mole of hydrochloric acid will neutralize 25 cent meter cubic of sodium carbonate solution
what is a volume 2 mole of hydrochloric acid will neutralize 25 cent meter cubic of 2 mole sodium carbonate solution
You should repost and make sure your post makes sense. What in the world is 25 cent meter cubic of 2 mol Na2CO3?
To calculate the volume of hydrochloric acid needed to neutralize a given volume of sodium carbonate solution, we can use the concept of mole-to-mole ratios from the balanced chemical equation.
Step 1: Write a balanced chemical equation for the reaction between hydrochloric acid (HCl) and sodium carbonate (Na2CO3):
2HCl + Na2CO3 -> 2NaCl + H2O + CO2
This equation shows that 2 moles of HCl react with 1 mole of Na2CO3 to produce 2 moles of NaCl, 1 mole of H2O, and 1 mole of CO2.
Step 2: Determine the molar ratio between HCl and Na2CO3. In this case, 2 moles of HCl react with 1 mole of Na2CO3.
Step 3: Convert the given volume of sodium carbonate solution (25 cm^3) into moles. To do this, we need the concentration of the sodium carbonate solution.
Step 4: Once you have the moles of Na2CO3, use the molar ratio from step 2 to calculate the moles of HCl required.
Step 5: Finally, convert the moles of HCl into volume by using the molar volume of the hydrochloric acid solution.
Note: Without the concentration of the sodium carbonate solution, it is not possible to give a specific answer.
For example, if the concentration of the sodium carbonate solution is 0.1 M (moles per liter), you can proceed with the calculations as follows:
Step 1: No change, as the balanced chemical equation remains the same.
Step 2: The molar ratio between HCl and Na2CO3 is 2:1.
Step 3: Convert the given volume of sodium carbonate solution (25 cm^3) into liters:
Volume = 25 cm^3 = 25/1000 L = 0.025 L
Step 4: Convert the volume of sodium carbonate solution into moles:
Moles of Na2CO3 = Concentration x Volume
Moles of Na2CO3 = (0.1 mol/L) x (0.025 L) = 0.0025 mol
Step 5: Use the molar ratio to find the moles of HCl:
Moles of HCl = 2 x Moles of Na2CO3
Moles of HCl = 2 x 0.0025 mol = 0.005 mol
Step 6: Finally, convert the moles of HCl into volume using the molar volume of hydrochloric acid solution. Assuming the concentration is 0.1 M, the molar volume is 0.1 L/mol:
Volume of HCl = Moles of HCl x Molar Volume
Volume of HCl = 0.005 mol x 0.1 L/mol = 0.0005 L or 0.5 cm^3
Therefore, 0.5 cm^3 of hydrochloric acid solution (0.1 M) will neutralize 25 cm^3 of 0.1 M sodium carbonate solution.