Final answer:
To find the volume of 0.2500 M cobalt(III) sulfate required to react completely with 5.00 g of sodium carbonate, use stoichiometry and the given balanced equation. The volume is approximately 0.5652 L or 565.2 mL.
Step-by-step explanation:
To find the volume of 0.2500 M cobalt(III) sulfate required to react completely with 5.00 g of sodium carbonate, we need to use stoichiometry.
The balanced equation for the reaction between cobalt(III) sulfate and sodium carbonate is:
Co2(SO4)3 + 3Na2CO3 → Co2(CO3)3 + 3Na2SO4
From the equation, we can see that the mole ratio of cobalt(III) sulfate to sodium carbonate is 1:3. This means that for every 1 mole of cobalt(III) sulfate, we need 3 moles of sodium carbonate.
First, we need to calculate the number of moles of sodium carbonate using its molar mass:
Number of moles = given mass / molar mass
Number of moles of sodium carbonate = 5.00 g / 105.99 g/mol = 0.0471 mol
Next, we can use the mole ratio to determine the number of moles of cobalt(III) sulfate:
Number of moles of cobalt(III) sulfate = (3/1) * 0.0471 mol = 0.1413 mol
Finally, we can use the molarity of cobalt(III) sulfate to find the volume:
Volume = number of moles / molarity
Volume of 0.2500 M cobalt(III) sulfate = 0.1413 mol / 0.2500 M = 0.5652 L or 565.2 mL