What volume of 4.5 M Acetic Acid is needed to neutralize 567.4 g sodium carbonate? ?
Volume acid in liters*(4.5)=molesNa2CO3
so calculate the moles of sodium carbonate in 567grams.
To find the volume of 4.5 M Acetic Acid needed to neutralize 567.4 g of sodium carbonate, we need to use the concept of stoichiometry.
First, we need to determine the balanced chemical equation between acetic acid (CH3COOH) and sodium carbonate (Na2CO3):
2CH3COOH + Na2CO3 -> 2CH3COONa + H2O + CO2
From the equation, we can see that it takes 2 moles of acetic acid to neutralize 1 mole of sodium carbonate.
Next, we need to convert the given mass of sodium carbonate (567.4 g) into moles. To do this, we use the molar mass of sodium carbonate (Na2CO3), which can be calculated by adding the atomic masses of each element:
Na: 2 x 22.99 g/mol = 45.98 g/mol
C: 1 x 12.01 g/mol = 12.01 g/mol
O: 3 x 16.00 g/mol = 48.00 g/mol
Molar mass of Na2CO3 = 45.98 g/mol + 12.01 g/mol + 48.00 g/mol = 105.99 g/mol
Now, we can calculate the number of moles of sodium carbonate by dividing the given mass by the molar mass:
Number of moles = Mass / Molar mass
Number of moles = 567.4 g / 105.99 g/mol ≈ 5.35 mol
Since the stoichiometric ratio between acetic acid and sodium carbonate is 2:1, it means that 2 moles of acetic acid react with 1 mole of sodium carbonate.
Therefore, the number of moles of acetic acid needed to neutralize the sodium carbonate is half the number of moles of sodium carbonate:
Number of moles of acetic acid = 5.35 mol / 2 = 2.67 mol
Finally, to find the volume of 4.5 M Acetic Acid needed, we can use the equation:
Molarity (M) = Moles / Volume (L)
4.5 M = 2.67 mol / Volume (L)
Rearranging the equation to solve for volume:
Volume (L) = Moles / Molarity
Volume (L) = 2.67 mol / 4.5 M ≈ 0.593 L
Therefore, the volume of 4.5 M Acetic Acid needed to neutralize 567.4 g of sodium carbonate is approximately 0.593 liters.